Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{x-1}{50}+\dfrac{x-2}{49}=\dfrac{x-3}{48}+\dfrac{x-4}{47}\)
\(\Rightarrow\dfrac{x-1}{50}-1+\dfrac{x-2}{49}-1=\dfrac{x-3}{48}-1+\dfrac{x-4}{47}-1\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}=\dfrac{x-51}{48}+\dfrac{x-51}{47}\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}-\dfrac{x-51}{48}-\dfrac{x-51}{47}=0\)
\(\Rightarrow\left(x-51\right)\left(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\right)=0\)
Vì \(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\ne0\) nên \(x-51=0\Rightarrow x=51\)
\(\dfrac{x+25}{6}+\dfrac{x+20}{11}+\dfrac{x+16}{15}+3=0\)
\(\Rightarrow\dfrac{x+25}{6}+1+\dfrac{x+20}{11}+1+\dfrac{x+16}{15}+1=0\)
\(\Rightarrow\dfrac{x+31}{6}+\dfrac{x+31}{11}+\dfrac{x+31}{15}=0\)
\(\Rightarrow\left(x+31\right)\left(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\ne0\) nên \(x+31=0\Rightarrow x=-31\)
\(\dfrac{x-15}{6}+\dfrac{x-10}{11}=\dfrac{x-3}{18}+\dfrac{x-7}{14}\)
\(\Rightarrow\dfrac{x-15}{6}-1+\dfrac{x-10}{11}-1=\dfrac{x-3}{18}-1+\dfrac{x-7}{14}-1\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}=\dfrac{x-21}{18}+\dfrac{x-21}{14}\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}-\dfrac{x-21}{18}-\dfrac{x-21}{14}=0\)
\(\Rightarrow\left(x-21\right)\left(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\ne0\) nên \(x-21=0\Rightarrow x=21\)
câu a) mình chịu (dùng kiến thức lớp 12 chắc làm đc haha)
b) gt ⇒ \(\frac{1}{6}.6^{x+2}-6^x=6^{14}-6^{13}\)
⇒ \(6^{x+1}-6^x=6^{14}-6^{13}\)
⇒ \(6^x\left(6-1\right)=6^{13}\left(6-1\right)\)
⇒ \(x=13\)
c) gt ⇒ \(\frac{1}{2}.2^{x+4}-2^x=2^{13}-2^{10}\)
⇒ \(2^{x+3}-2^x=2^{13}-2^{10}\)
⇒ \(2^x\left(2^3-1\right)=2^{10}\left(2^3-1\right)\)
⇒ \(x=10\)
d) gt ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
⇒ \(x=13\)
Các câu dễ tự làm :v
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
\(a,x^2=16\)
\(x^2=4^2=\left(-4\right)^2\)
\(x=2\) hoặc \(x=-2\)
\(b,x^3=-8\)
\(x^3=\left(-2\right)^3\)
\(x=-2\)
\(c,\left(x+2\right)^2=4\)
\(\left(x+2\right)^2=2^2=\left(-2\right)^2\)
\(x+2=2\Rightarrow x=0\) hoặc \(x+2=-2\Rightarrow x=-4\)
\(d,\left(1-x\right)^3=1\)
\(1-x=1\)
\(x=0\)
e,phần này mk chưa nghĩ ra,sorry bn nha!
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
3, Tìm x, biết
\(d,\dfrac{-16}{x}=\dfrac{x}{-4}=>x^2=\left(-16\right).\left(-4\right)=>x^2=64\)
\(=>x=8\) hay \(x=-8\)
\(e,\dfrac{x}{-2}=\dfrac{\dfrac{8}{25}}{-x}=>-x^2=-2.\dfrac{8}{5}=\dfrac{-16}{25}\)
\(=>-x^2=0,64=>x=0,8\)
\(g,\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(=>x^2=\left(-15\right).\left(-60\right)\)\(=>x^2=900=>x=30\) hay \(x=-30\)
d) \(\dfrac{-16}{x}=\dfrac{x}{-4}\)
= 16 . 4 = x.x
= 64 = \(x^2\)
= \(8^2=x^2\)
vậy x = 8
e)\(\dfrac{x}{-2}=\dfrac{8}{\dfrac{25}{-x}}\)
= -2 . \(\dfrac{8}{25}\) = -x . x
= -0,64 = \(-x^2\)
= 0,64 = \(x^2\)
0,8\(^2=x^2\)
vậy x = 0,8
g) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
= -15 . -60 = x.x
= 900 = \(x^2\)
30 \(^2=x^2\)
vậy x = 30
1: =>1/3:x=3/5-2/3=9/15-10/15=-1/15
=>x=-1/3:1/15=5
2: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)
=>x*2/3=-1
=>x=-3/2
3: \(\Leftrightarrow\dfrac{8}{3}:x=\dfrac{25}{12}:\dfrac{-3}{50}=\dfrac{25}{12}\cdot\dfrac{-50}{3}\)
hay x=-48/625
9: =>x=-2*3/1,5=-4
8: =>2/3:x=5/2:-3/10=5/2*(-10)/3=-50/6=-25/3
=>x=-2/3:25/3=-2/3*3/25=-2/25
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
<=> \(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
=> x+1=0
<=> x=-1
b) \(\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
<=> \(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
đến đây tương tự a
a) Ta có:
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(Vì:\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
Vậy....
b)Sửa lại đề nha
Ta có:
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Leftrightarrow\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
\(\Leftrightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
Lý giải tương tự câu a và kết luận nha
a) Ta có : \(x - 2xy + y - 3 = 0\)
\(\Rightarrow-2xy+x+y=3\)
\(\Rightarrow-2.\left(-2xy+x+y\right)=-2.3\)
\(\Rightarrow4xy-2x-2y=-6\)
\(\Rightarrow4xy-2x-2y+1=-6+1\)
\(\Rightarrow2x.\left(2y-1\right).\left(2y-1\right)=-5\)
\(\Rightarrow\left(2y-1\right).\left(2x-1\right)=-5=1.\left(-5\right)=-5.1=\left(-1\right).5=5.\left(-1\right)\)
Tự lập bảng đi -.-
|
Nhân từng vế bất đẳng thức ta được : (xyz)2 = 36xyz + Nếu một trong các số x,y,z bằng 0 thì 2 số còn lại cũng bằng 0 + Nếu cả 3 số x,y,z khác 0 thì chia 2 vế cho xyz ta được xyz = 36 + Từ xyz =36 và xy = z ta được z2 = 36 nên z = 6; z = -6 + Từ xyz =36 và yz = 4x ta được 4x2 = 36 nên x = 3; x = -3 + Từ xyz =36 và ta được 9y2 = 36 nên y = 2; y = -2 - Nếu z = 6 thì x và y cùng dấu nên x = 3, y = 2 hoặc x = -3 , y = -2 - Nếu z = -6 thì x và y trái dấu nên x = 3 ; y = -2 hoặc x = -3; y=2 |
Vậy có 5 bộ số (x, y, z) thoã mãn: (0,0,0); (3,2,6);(-3,-2,6);(3,-2,-6);(-3,2.-6)
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)
\(\Leftrightarrow\dfrac{1680.\left(x-3\right)+1560.\left(x-3\right)-1456.\left(x-3\right)-1365.\left(x-3\right)}{21840}=0\)
\(\Leftrightarrow\left(x-3\right).\left(1680+1560-1456-1365\right)=0\)
\(\Leftrightarrow\left(x-3\right).419=0\)
\(\Leftrightarrow419x=1257\)
\(\Leftrightarrow x=3\)
Lời giải:
\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\((x-3)\left(\frac{1}{13}+\frac{1}{14}\right)=(x-3)\left(\frac{1}{15}+\frac{1}{16}\right)\)
\((x-3)\left[\left(\frac{1}{13}+\frac{1}{14}\right)-\left(\frac{1}{15}+\frac{1}{16}\right)\right]=0\)
Ta thấy:
\(\frac{1}{13}>\frac{1}{15}; \frac{1}{14}>\frac{1}{16}\Rightarrow \frac{1}{13}+\frac{1}{14}> \frac{1}{15}+\frac{1}{16}\)
Do đó biểu thức trong ngoặc vuông lớn hơn $0$ hay khác $0$
$\Rightarrow x-3=0$
$\Leftrightarrow x=3$