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\(\frac{x-2}{4}=\frac{-9}{2-x}\)
\(\Rightarrow\frac{x-2}{4}=\frac{9}{x-2}\)
\(\Rightarrow\left(x-2\right)^2=36\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}}\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\Rightarrow3\left(2x+1\right)=\left(x+2\right)5\)
\(\Rightarrow6x+3=5x+10\)
\(\Rightarrow6x-5x=10-3\)
\(\Rightarrow x=7\)
c;giống câu trên :v
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Ta có:3(x+1)+24=16-(12-2x)
3x+3+24=16-12+2x
3x+27=4+2x
3x+27-2x=4
3x-2x+27=4
x+27=4
x=-23 Vậy x=-23
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\(\frac{5}{2}x+\frac{1}{2}x=x+400\%\)
\(\Rightarrow\left(\frac{5}{2}+\frac{1}{2}\right)x=x+4\)
\(\Rightarrow\frac{6}{2}x=x+4\)
\(\Rightarrow3x=x+4\)
\(\Rightarrow3x-x=4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
Chúc bạn học tốt !!!
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a) \(|x+1|=3\)
\(\Rightarrow x+1=\pm3\)
+) \(x+1=3\) +) \(x+1=-3\)
\(\Rightarrow x=2\) \(\Rightarrow x=-4\)
Vậy \(x\in\left\{2;-4\right\}\)
b) \(3^2x+2^4=5^2\)
\(9x+16=25\)
\(9x=25-16\)
\(9x=9\)
\(x=1\)
c) \(\frac{4+x}{7+y}=\frac{4}{7}\)
\(\Rightarrow\left(4+x\right).7=\left(7+y\right).4\)
\(\Rightarrow28+7x=28+4y\)
\(\Rightarrow7x=4y\)
Mà \(\left(7,4\right)=1\) và \(x+y=11\)
Vậy \(x=4;y=7\)
a) Ta có: \(\left|x+1\right|=3\)
\(\Rightarrow x+1=\pm3\)
Nếu x + 1 = 3 => x = 2
Nếu x + 1 = -3 => x = -4
Vậy x = {2;-4}
b) \(3^2x+2^4=5^2\)
\(\Rightarrow9x+16=25\)
\(\Rightarrow9x=9\)
\(\Rightarrow x=1\)
Vậy x = 1
c) \(\frac{4+x}{7+x}=\frac{4}{7}\)
\(\Rightarrow7\left(4+x\right)=4\left(7+x\right)\)
\(\Rightarrow28+7x=28+4x\)
\(\Rightarrow7x-4x=0\)
\(\Rightarrow x=0\)
Vậy x = 0
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4x-2x+x-27:9=33
2x+x-27:9=33
3x-27:9=33
3x-27=33×9=297
3x=297+27=324
x=324÷3=108
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