Giải phương trình:
\(\dfrac{x-1}{13}-\dfrac{2x-13}{15}=\dfrac{3x-15}{27}-\dfrac{4x-27}{29}\)
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11 tháng 3 2021
1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)
\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)
\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))
\(\Leftrightarrow x=-36\).
Vậy nghiệm của pt là x = -36.
17 tháng 7 2024
2) x(x+1)(x+2)(x+3)= 24
⇔ x.(x+3) . (x+2).(x+1) = 24
⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24
Đặt \(x^2\)+ 3x = b
⇒ b . (b+2)= 24
Hay: \(b^2\) +2b = 24
⇔\(b^2\) + 2b + 1 = 25
⇔\(\left(b+1\right)^2\)= 25
+ Xét b+1 = 5 ⇒ b=4 ⇒ \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0
⇒(x-1)(x+4)=0⇒x=1 và x=-4
+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0
⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))2 = - \(\dfrac{15}{4}\) Hay ( \(x^2\) +\(\dfrac{3}{2}\) )2= -\(\dfrac{15}{4}\) (vô lí)
⇒x= 1 và x= 4
ND
30 tháng 5 2018
\(\dfrac{x-1}{13}-\dfrac{2x-13}{15}=\dfrac{3x-15}{27}-\dfrac{2x-27}{29}\)
\(\Leftrightarrow\dfrac{x-1}{13}-1-\dfrac{2x-13}{15}-1=\dfrac{3x-15}{27}-1-\dfrac{2x-27}{29}-1\)
\(\Leftrightarrow\dfrac{x-1-13}{13}-\dfrac{2x-13-15}{15}=\dfrac{3x-15-27}{27}-\dfrac{4x-27-29}{29}\)
\(\Leftrightarrow\dfrac{x-14}{13}-\dfrac{2x-24}{15}=\dfrac{3x-42}{27}-\dfrac{4x-56}{29}\)
\(\Leftrightarrow\dfrac{x-14}{13}-\dfrac{2\left(x-14\right)}{15}-\dfrac{3\left(x-14\right)}{27}-\dfrac{4\left(x-14\right)}{29}=0\)
\(\Leftrightarrow\left(x-14\right)\left(\dfrac{1}{13}-\dfrac{2}{15}-\dfrac{3}{27}-\dfrac{4}{29}\right)=0\)
\(\Leftrightarrow x-14=0\) ( Vì: \(\dfrac{1}{13}-\dfrac{2}{15}-\dfrac{3}{27}-\dfrac{4}{29}\ne0\))
\(\Leftrightarrow x=14\)
2 tháng 3 2018
\(\text{a) }\left|2-5x\right|=\left|3x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}2-5x=3x+1\\2-5x=-3x-1\end{matrix}\right. \Leftrightarrow\left[{}\begin{matrix}-5x-3x=1-2\\-5x+3x=-1-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-8x=-1\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{\dfrac{1}{8};\dfrac{3}{2}\right\}\)
\(\text{b) }\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\)
ĐXKĐ của phương trình \(:x\ne\pm5\)
\(\text{Ta có }:\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}+\dfrac{15}{2\left(25-x^2\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x+5\right)\left(x-5\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{9\left(x+5\right)}{12\left(x+5\right)\left(x-5\right)}-\dfrac{90}{12\left(x+5\right)\left(x-5\right)}+\dfrac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}=0\\ \Rightarrow9x+45-90+14x-70=0\\ \Leftrightarrow23x=115\\ \Leftrightarrow x=5\left(KTM\right)\)
Vậy phương trình vô nghiệm
\(\text{c) }\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\\ \Leftrightarrow\left(\dfrac{x+29}{31}+1\right)-\left(\dfrac{x+27}{33}+1\right)=\left(\dfrac{x+17}{43}+1\right)-\left(\dfrac{x+15}{45}+1\right)\\ \Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}-\dfrac{x+60}{43}+\dfrac{x+60}{45}=0\\ \Leftrightarrow\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\\ \Leftrightarrow x+60=0\left(\text{Vì }\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\right)\\ \Leftrightarrow x=-60\)
Vậy \(x=-60\) là nghiệm của phương trình
11 tháng 7 2017
Giải phương trình
\(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\)\(\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\)
\(\Leftrightarrow\)\(\dfrac{x+2}{13}+\dfrac{13}{13}+\dfrac{2x+45}{15}-\dfrac{15}{15}=\dfrac{3x+8}{37}+\dfrac{37}{37}+\dfrac{4x+69}{9}-\dfrac{9}{9}\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2x+30}{15}=\dfrac{3x+45}{37}+\dfrac{4x+60}{9}\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}=\dfrac{3\left(x+15\right)}{37}+\dfrac{4\left(x+15\right)}{9}\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}\right)=\left(x+15\right)\left(\dfrac{3}{37}+\dfrac{4}{9}\right)\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}\right)-\left(x+15\right)\left(\dfrac{3}{37}+\dfrac{4}{9}\right)=0\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=0\\\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-15\\\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\ne0\end{matrix}\right.\)
Do đó: \(x=-15\)
Vậy \(S=\left\{-15\right\}\)
AT
14 tháng 6 2017
\(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\)\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}=\dfrac{3\left(x+15\right)}{37}+\dfrac{4\left(x+15\right)}{9}\)\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}-\dfrac{3\left(x+15\right)}{37}-\dfrac{4\left(x+15\right)}{9}=0\)\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}+\dfrac{4}{9}\right)=0\)
\(\Leftrightarrow x+15=0\)
\(\Leftrightarrow x=-15\)
Vậy x = -15.
14 tháng 6 2017
\(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\\ \Leftrightarrow\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\\ \Leftrightarrow\dfrac{x+15}{13}+\dfrac{2x+30}{15}=\dfrac{3x+45}{37}+\dfrac{4x+60}{9}\)
\(\Leftrightarrow\left(x+15\right)\dfrac{1}{13}+\left(x+15\right)\dfrac{2}{15}=\left(x+15\right)\dfrac{3}{37}+\left(x+15\right)\dfrac{4}{9}\\ \Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
vì:\(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\ne0\) nên:
x+15=0 =>x=-15
vậy phương trình có tập nghiệm là S={-15}
TL
giải phương trình sau
\(\dfrac{\text{x+1}}{11}-\dfrac{2x-5}{15}=\dfrac{3x-47}{17}-\dfrac{4x-59}{19}\)
1
AH
Akai Haruma
Giáo viên
14 tháng 8 2021
Lời giải:
PT \(\Leftrightarrow (\frac{x+1}{11}-1)-(\frac{2x-5}{15}-1)=(\frac{3x-47}{17}+1)-(\frac{4x-59}{19}+1)\)
\(\Leftrightarrow \frac{x-10}{11}-\frac{2(x-10)}{15}=\frac{3(x-10)}{17}-\frac{4(x-10)}{19}\)
\(\Leftrightarrow (x-10)(\frac{1}{11}+\frac{4}{19}-\frac{2}{15}-\frac{3}{17})=0\)
\(\Leftrightarrow x-10=0\Leftrightarrow x=10\)
28 tháng 9 2021
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
19 tháng 2 2022
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
\(\dfrac{x-1}{13}-\dfrac{2x-13}{15}=\dfrac{3x-15}{27}-\dfrac{4x-27}{29}\)
=>\(\left(\dfrac{x-1}{13}-1\right)-\left(\dfrac{2x-13}{15}-1\right)=\left(\dfrac{3x-15}{27}-1\right)-\left(\dfrac{4x-27}{29}-1\right)\)
=>\(\dfrac{x-14}{13}-\dfrac{2x-28}{15}-\dfrac{3x-42}{27}+\dfrac{4x-56}{29}=0\)
=>\(\left(x-14\right)\left(\dfrac{1}{13}-\dfrac{2}{15}-\dfrac{3}{27}+\dfrac{4}{29}\right)=0\)
=>x-14=0
=>x=14