\(n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right)\)

\(n_{AgNO_3}=2.0,16=0,32\left(mol\right)\)

PTHH:

\(Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\)

0,125       0,25           0,125           0,25

Số mol Cu phản ứng :

\(n_{Cu\left(pư\right)}=\dfrac{51-32}{2.108-64}=0,125\left(mol\right)\)

a,\(C_{M\left(Cu\left(NO_3\right)_2\right)}=\dfrac{0,125}{2}=\dfrac{1}{16}\left(M\right)\)

\(C_{M\left(AgNO_3dư\right)}=\dfrac{0,32-0,25}{2}=\dfrac{7}{200}\left(M\right)\)

Câu b để mình suy nghĩ sau:)

a, BaO, CO2, N2O5, Na2O, P2O5

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(CO_2+H_2O⇌H_2CO_3\)

\(N_2O_5+H_2O\rightarrow2HNO_3\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

b, BaO, Na2O, ZnO, CuO, Fe2O3

\(BaO+H_2SO_4\rightarrow BaSO_4\downarrow+H_2O\)

\(Na_2O+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

c, CO2, N2O5, P2O5

\(CO_2+K_2O\rightarrow K_2CO_3\)

\(N_2O_5+K_2O\rightarrow2KNO_3\)

\(3K_2O+P_2O_5\rightarrow2K_3PO_4\)

d, CO2, N2O5, P2O5 

\(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)

\(N_2O_5+2KOH\rightarrow2KNO_3+H_2O\)

\(P_2O_5+6KOH\rightarrow2K_3PO_4+3H_2O\)

e, BaO, Na2O, ZnO,CuO

\(BaO+CO_2\rightarrow BaCO_3\)

\(Na_2O+CO_2\rightarrow Na_2CO_3\)

\(ZnO+CO_2\rightarrow ZnCO_3\)

\(CuO+CO_2\rightarrow CuCO_3\)

Bài 1:

e: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)

=>\(\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)

=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)

=>\(\left(x+1+x-1\right)\left(x+1-x+1\right)=16\)

=>4x=16

=>x=4(nhận)

f: ĐKXĐ: \(x\notin\left\{1-1\right\}\)

\(\left(1-\dfrac{x-1}{x+1}\right)\left(x+2\right)=\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\)

=>\(\dfrac{x+1-x+1}{\left(x+1\right)}\left(x+2\right)=\dfrac{\left(x+1\right)^2+\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)

=>\(\dfrac{2\left(x+2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}\)

=>\(2\left(x+2\right)\left(x-1\right)=2\left(x^2+1\right)\)

=>\(\left(x+2\right)\left(x-1\right)=x^2+1\)

=>\(x^2+x-2=x^2+1\)

=>x-2=1

=>x=3(nhận)

a: ĐKXĐ: \(x\notin\left\{0;-1;4\right\}\)

\(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\)

=>\(\dfrac{11}{x}=\dfrac{9\left(x-4\right)+2\left(x+1\right)}{\left(x+1\right)\left(x-4\right)}\)

=>\(\dfrac{11}{x}=\dfrac{11x-34}{x^2-3x-4}\)

=>\(11\left(x^2-3x-4\right)=x\left(11x-34\right)\)

=>\(11x^2-33x-44=11x^2-34x\)

=>-33x-44=-34x

=>-33x+34x=44

=>x=44(nhận)

b: ĐKXĐ: \(x\ne4\)

\(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)

=>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

=>\(\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)

=>28-6(x+2)=-9-5(x-4)

=>28-6x-12=-9-5x+20

=>-6x+16=-5x+11

=>-6x+5x=11-16

=>-x=-5

=>x=5(nhận)

c: ĐKXĐ: \(x\notin\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)

\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)

=>\(\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

=>\(\left(1-3x\right)^2-\left(1+3x\right)^2=12\)

=>\(9x^2-6x+1-9x^2-6x-1=12\)

=>-12x=12

=>x=-1(nhận)

d: ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)

\(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{x-5}{2x^2+10x}\)

=>\(\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{2x\left(x+5\right)}\)

=>\(\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}\)

=>\(2\left(x+5\right)^2-x\left(x+25\right)=\left(x-5\right)^2\)

=>\(2x^2+20x+50-x^2-25x=x^2-10x+25\)

=>-5x+50=-10x+25

=>5x=-25

=>x=-5(loại)

Bài 2:

a: ĐKXĐ: \(x\notin\left\{2;5\right\}\)

\(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)

=>\(\dfrac{6x+1}{\left(x-2\right)\left(x-5\right)}+\dfrac{5\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}\)

=>6x+1+5x-25=3x-6

=>11x-24=3x-6

=>8x=18

=>x=9/4(nhận)

b: ĐKXĐ: \(x\notin\left\{0;2;-2\right\}\)

\(\dfrac{2}{x^2-4}-\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x-4}{x\left(x+2\right)}=0\)

=>\(\dfrac{2x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)

=>2x-(x-1)(x+2)+(x-4)(x-2)=0

=>\(2x-\left(x^2+x-2\right)+x^2-6x+8=0\)

=>\(x^2-4x+8-x^2-x+2=0\)

=>-5x+10=0

=>x=2(loại)

c: ĐKXĐ: \(x\notin\left\{3;-1\right\}\)

\(\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\)

=>\(\dfrac{-1}{x-3}-\dfrac{1}{x+1}-\dfrac{x}{x-3}+\dfrac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)

=>\(\dfrac{\left(-1-x\right)\left(x+1\right)-x+3}{\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)

=>-(x+1)^2-x+3+(x-1)²=0

=>\(-x^2-2x-1-x+3+x^2-2x+1=0\)

=>-5x+3=0

=>\(x=\dfrac{3}{5}\left(nhận\right)\)

d: ĐKXĐ: \(x\notin\left\{2;-3\right\}\)

\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)

=>\(\dfrac{x+3-6\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{-5}{\left(x+3\right)\left(x-2\right)}\)

=>x+3-6(x-2)=-5

=>x+3-6x+12+5=0

=>-5x+20=0

=>x=4(nhận)

e: ĐKXĐ: x<>-2

\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)

=>\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5}{x^2-2x+4}=0\)

=>\(\dfrac{2\left(x^2-2x+4\right)-2x^2-16-5x-10}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)

=>\(2x^2-4x+8-2x^2-5x-26=0\)

=>-9x-18=0

=>x=-2(loại)

f: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)

=>\(\dfrac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

=>\(\dfrac{2}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{2\left(x+2\right)^2}{\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

=>2(x^2-1)=2(x+2)^2

=>\(x^2-1=\left(x+2\right)^2\)

=>\(x^2+4x+4-x^2+1=0\)

=>4x+5=0

=>\(x=-\dfrac{5}{4}\left(nhận\right)\)

Bài 3:

c:

=>\(\dfrac{x}{x-1}+\dfrac{x}{x-2}+\dfrac{x}{x-3}=\dfrac{3x-12}{x-6}\)

=>

ĐKXĐ: \(x\notin\left\{1;2;\dfrac{3\pm\sqrt{7}}{2}\right\}\)

 \(\dfrac{4}{x^2-3x+2}-\dfrac{3}{2x^2-6x+1}+1=0\)

=>\(\dfrac{4\left(2x^2-6x+1\right)-3\left(x^2-3x+2\right)}{\left(x^2-3x+2\right)\left(2x^2-6x+1\right)}=-1\)

=>\(8x^2-24x+4-3x^2+9x-6=-\left(x^2-3x+2\right)\left[2\cdot\left(x^2-3x\right)+1\right]\)

=>\(5x^2-15x-2=-\left[2\left(x^2-3x\right)^2+5\left(x^2-3x\right)+2\right]\)

=>\(5\left(x^2-3x\right)-2+2\left(x^2-3x\right)^2+5\left(x^2-3x\right)+2=0\)

=>\(2\left(x^2-3x\right)^2+10\left(x^2-3x\right)=0\)

=>\(\left(x^2-3x\right)^2+5\left(x^2-3x\right)=0\)

=>\(\left(x^2-3x\right)\left(x^2-3x+5\right)=0\)

mà \(x^2-3x+5=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}>0\forall x\)

nên x(x-3)=0

=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

a:

ĐKXĐ: \(x\notin\left\{8;9;10;11\right\}\)

 \(\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\)

=>\(\left(\dfrac{8}{x-8}+1\right)+\left(\dfrac{11}{x-11}+1\right)=\left(\dfrac{9}{x-9}+1\right)+\left(\dfrac{10}{x-10}+1\right)\)

=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)

=>\(x\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}\right)=0\)

=>x=0(nhận)

b:

ĐKXĐ: \(x\notin\left\{3;4;5;6\right\}\)

 \(\dfrac{x}{x-3}-\dfrac{x}{x-5}=\dfrac{x}{x-4}-\dfrac{x}{x-6}\)

=>\(\dfrac{x\left(x-5\right)-x\left(x-3\right)}{\left(x-3\right)\left(x-5\right)}=\dfrac{x\left(x-6\right)-x\left(x-4\right)}{\left(x-4\right)\left(x-6\right)}\)

=>\(\dfrac{-2x}{\left(x-3\right)\left(x-5\right)}=\dfrac{-2x}{\left(x-4\right)\left(x-6\right)}\)

=>\(x\left(\dfrac{1}{\left(x-3\right)\left(x-5\right)}-\dfrac{1}{\left(x-4\right)\left(x-6\right)}\right)=0\)

=>\(x\cdot\dfrac{\left(x-4\right)\left(x-6\right)-\left(x-3\right)\left(x-5\right)}{\left(x-3\right)\left(x-5\right)\left(x-4\right)\left(x-6\right)}=0\)

=>\(x\left(x^2-10x+24-x^2+8x-15\right)=0\)

=>x(-2x+9)=0

=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=\dfrac{9}{2}\left(nhận\right)\end{matrix}\right.\)

\(A=\left(\dfrac{x+1}{x^3-1}-\dfrac{1}{x-1}\right)\left(\dfrac{x+2}{x-1}-\dfrac{1}{x}\right)\left(x\ne1;0\right)\\ =\left[\dfrac{x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\left[\dfrac{x\left(x+2\right)}{x\left(x-1\right)}-\dfrac{x-1}{x\left(x-1\right)}\right]\\ =\dfrac{x+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+2x-x+1}{x\left(x-1\right)}\\ =\dfrac{-x^2}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x\left(x-1\right)}\\ =\dfrac{-x}{\left(x-1\right)^2}\\ =\dfrac{-x}{x^2-2x+1}\)

ĐKXĐ: \(x\notin\left\{1;0\right\}\)

\(A=\left(\dfrac{x+1}{x^3-1}-\dfrac{1}{x-1}\right)\left(\dfrac{x+2}{x-1}-\dfrac{1}{x}\right)\)

\(=\left(\dfrac{x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right)\cdot\left(\dfrac{x\left(x+2\right)-x+1}{x\left(x-1\right)}\right)\)

\(=\dfrac{x+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x\left(x-1\right)}\)

\(=\dfrac{-x^2}{\left(x-1\right)\cdot x\left(x-1\right)}=\dfrac{-x}{\left(x-1\right)^2}\)

a: Xét (O) có

CM,CA là các tiếp tuyến

Do đó: CM=CA và OC là phân giác của góc MOA

Xét (O) có

DM,DB là các tiếp tuyến

Do đó: DM=DB và OD là phân giác của góc MOB

AC+BD

=CM+MD

=CD
b: \(\widehat{COD}=\widehat{COM}+\widehat{DOM}=\dfrac{1}{2}\cdot\widehat{MOA}+\dfrac{1}{2}\cdot\widehat{MOB}\)

\(=\dfrac{1}{2}\left(\widehat{MOA}+\widehat{MOB}\right)=\dfrac{1}{2}\cdot\widehat{AOB}=90^0\)

=>ΔCOD vuông tại O

c: Xét ΔCOD vuông tại O có OM là đường cao

nên \(OM^2=MC\cdot MD\)

giúp tôi ý d với bạn ơi

Bài 4:

d: 

ĐKXĐ: \(x\notin\left\{1;-1;2;-2\right\}\)

\(\dfrac{x+4}{x-1}+\dfrac{x-4}{x+1}=\dfrac{x+8}{x-2}+\dfrac{x-8}{x+2}+6\)

=>\(\dfrac{\left(x+4\right)\left(x+1\right)+\left(x-4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+8\right)\left(x+2\right)+\left(x-8\right)\left(x-2\right)+6\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>\(\dfrac{2x^2+8}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x^2+32+6x^2-24}{\left(x-2\right)\left(x+2\right)}\)

=>\(\dfrac{2x^2+8}{x^2-1}=\dfrac{8x^2+8}{x^2-4}\)

=>\(\left(2x^2+8\right)\left(x^2-4\right)=\left(8x^2+8\right)\left(x^2-1\right)\)

=>\(2x^4-32=8x^4-8\)

=>\(-6x^4=24\)

=>\(x^4=-4\left(loại\right)\)

Vậy: Phương trình vô nghiệm

c:

ĐKXĐ: \(x\notin\left\{-1;-3;-8;-10\right\}\)

 \(\dfrac{2}{x^2+4x+3}+\dfrac{5}{x^2+11x+24}+\dfrac{2}{x^2+18x+80}=\dfrac{9}{52}\)

=>\(\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{52}\)

=>\(\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{52}\)

=>\(\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{52}\)

=>\(\dfrac{9}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{52}\)

=>(x+1)(x+10)=52

=>\(x^2+11x-42=0\)

=>(x+14)(x-3)=0

=>\(\left[{}\begin{matrix}x=-14\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b: 

ĐXKĐ: \(x\notin\left\{-2;-3;-4;-5;-6\right\}\)\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)

=>\(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

=>\(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

=>\(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

=>\(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)

=>(x+2)(x+6)=32

=>\(x^2+8x-20=0\)

=>(x+10)(x-2)=0

=>\(\left[{}\begin{matrix}x=-10\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

a: \(\dfrac{x^2}{x^2+2x+2}+\dfrac{x^2}{x^2-2x+2}-\dfrac{4x^2-20}{x^4+4}=\dfrac{322}{65}\)

=>\(\dfrac{x^2\left(x^2-2x+2\right)+x^2\left(x^2+2x+2\right)-4x^2+20}{\left(x^2+2x+2\right)\left(x^2-2x+2\right)}=\dfrac{322}{65}\)

=>\(\dfrac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)

=>\(\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)

=>\(322\left(x^4+4\right)=65\left(2x^4+20\right)\)

=>\(322x^4+1288-130x^4-1300=0\)

=>\(192x^4=12\)

=>\(x^4=\dfrac{1}{16}\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\left(nhận\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)

a: \(\sqrt{14-6\sqrt{5}}=\sqrt{9-2\cdot3\cdot\sqrt{5}+5}\)

\(=\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

b: \(\sqrt{7-4\sqrt{7}+4}\)

\(=\sqrt{\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot2+2^2}\)

\(=\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|=\sqrt{7}-2\)

ĐKXĐ: \(x\notin\left\{1;7\right\}\)

\(\dfrac{x-8}{x-7}=8+\dfrac{1}{1-x}\)

=>\(\dfrac{x-8}{x-7}=\dfrac{8-8x+1}{1-x}\)

=>\(\dfrac{x-8}{x-7}=\dfrac{-8x+9}{1-x}\)

=>\(\dfrac{x-8}{x-7}=\dfrac{8x-9}{x-1}\)

=>\(\left(8x-9\right)\left(x-7\right)=\left(x-8\right)\left(x-1\right)\)

=>\(8x^2-65x+63-x^2+9x-8=0\)

=>\(7x^2-56x+55=0\)

\(\text{Δ}=\left(-56\right)^2-4\cdot7\cdot55=1596>0\)

=>Phương trình có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}x=\dfrac{56-2\sqrt{399}}{2\cdot7}=\dfrac{28-\sqrt{399}}{7}\left(nhận\right)\\x=\dfrac{28+\sqrt{399}}{7}\left(nhận\right)\end{matrix}\right.\)