\(\frac{32-x}{7}=\frac{x-42}{9}\)

=\(\frac{\left(32-x\right)9}{63}=\frac{\left(x-42\right)7}{63}\)

\(\Rightarrow\)\(\left(32-x\right)9=\left(x-42\right)7\)

=\(288-x9=x7-294\)

=\(288+294=x9+x7\)

=\(x=-36\frac{6}{16}\)

=\(x\times16=-582\)

\(x=-582\div16\)

a,\(\frac{32-x}{7}=\frac{x-42}{9}\)

\(\Leftrightarrow9\left(32-x\right)=7\left(x-42\right)\)

\(\Leftrightarrow288-9x-7x-294=0\)

\(\Leftrightarrow9x+7x=288-294\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

b. \(\left(2x-1\right)^2+\left|x+3\right|=0\)

\(\Leftrightarrow\left|x+3\right|=-4x^2+4x-1\)

\(\left|x+3\right|=x+3\)khi \(x+3\ge0\)hay \(x\ge-3\)

\(\left|x+3\right|=-\left(x+3\right)\)khi \(x+3< 0\)hay \(x< -3\)

với \(x\ge-3\Rightarrow x+3=-4x^2+4x-1\)

\(\Leftrightarrow4x^2-4x+1+x+3=0\)

\(\Leftrightarrow4x^2-3x+4=0\)\(\Leftrightarrow\)vô nghiệm

với \(x< -3\)\(\Rightarrow-x-3=-4x+4-1\)

\(\Leftrightarrow4x^2-4x+1-x-3=0\)

\(\Leftrightarrow4x^2-5x-2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{57}}{8}\left(tm\right)\\x=\frac{5-\sqrt{57}}{8}\left(L\right)\end{cases}}\)

1.\(\frac{x+1}{x-2}=\frac{3}{4}\)

\(\Leftrightarrow\left(x+1\right).4=\left(x-2\right).3\)

\(\Leftrightarrow4x+4=3x-6\)

<=>4x-3x=-6-4

<=>x=-10

2.\(\frac{52}{2x-1}=\frac{13}{30}\)

<=>52.30=(2x-1).13

<=>1560=26x-13

<=>-26x=-13-1560

<=>-26x=-1573

<=>x=60,5

3.\(\frac{2x-3}{x+1}=\frac{4}{7}\)

<=>(2x-3).7=(x+1).4

<=>14x-21=4x+4

<=>14x-4x=4+21

<=>10x=25

<=>x=2,5

4.\(\frac{2x+3}{42}=\frac{3x-1}{32}\)

<=>(2x+3).32=42(3x-1)

<=>64x+96=126x-42

<=>64x-126x=-42-96

<=>-62x=-138

<=>x=69/31

cho mk hỏi 

10x=25 số 10 ở đâu vậy bn 

a,    x=10

b,   x= 2

A/\(\left(2,8x-32\right):\frac{2}{3}=-90\)

\(\left(\frac{28}{10}x-32\right)=\frac{-90}{1}.\frac{2}{3}\)

\(\left(\frac{14}{5}x-32\right)=\frac{-30}{1}.\frac{2}{1}\)

\(\left(\frac{14}{5}x-32\right)=-60\)

\(\frac{14}{5}x=-60+32\)

\(\frac{14}{5}x=-28\)

\(x=\frac{-28}{1}:\frac{14}{5}\)

\(x=\frac{-28}{1}.\frac{5}{14}\)

\(x=\frac{-2}{1}.\frac{5}{1}=-10\)

B/\(\left(4,5-2x\right).1\frac{4}{7}=\frac{11}{14}\)

\(\left(\frac{45}{10}-2x\right).\frac{11}{7}=\frac{11}{14}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}:\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}.\frac{7}{11}\)

\(\left(\frac{9}{2}-2x\right)=\frac{1}{2}.\frac{1}{1}=\frac{1}{2}\)

\(2x=\frac{9}{2}-\frac{1}{2}\)

\(2x=\frac{8}{2}\)

\(x=\frac{8}{2}:\frac{2}{1}=\frac{8}{2}.\frac{1}{2}\)

\(x=\frac{4}{2}.\frac{1}{1}=\frac{4}{2}=2\)

\(1,x^3-3x^2=0\)

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

\(\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=3\left(TM\right)\end{cases}}}\)

\(2,3x^3-48x=0\)

\(3x\left(x^2-16\right)=0\)

\(\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x^2=16\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=\pm4\left(TM\right)\end{cases}}}}\)

\(3,5x\left(x-1\right)=x-1\)

\(5x^2-5x=x-1\)

\(5x^2-6x+1=0\)

\(5x^2-5x-x+1=0\)

\(5x\left(x-1\right)-\left(x-1\right)=0\)

\(\left(5x-1\right)\left(x-1\right)=0\)

\(\orbr{\begin{cases}5x-1=0\\x-1=0\end{cases}\orbr{\begin{cases}x=\frac{1}{5}\left(TM\right)\\x=1\left(TM\right)\end{cases}}}\)

\(4,2\left(x+5\right)-x^2-5x=0\)

\(2x+10-x^2-5x=0\)

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

\(-x^2-5x+2x+10=0\)

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

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

\(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\orbr{\begin{cases}x=-5\left(TM\right)\\x=2\left(TM\right)\end{cases}}}\)

\(5,2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(2x^2-10x-3x-2x^2=26\)

\(-13x-26=0\)

\(-13\left(x+2\right)=0\)

\(x=-2\left(TM\right)\)

Trả lời:

1, \(x^3-3x^2=0\)

\(\Leftrightarrow x^2\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)

Vậy x = 0; x = 3 là nghiệm của pt.

2, \(3x^3-48x=0\)

\(\Leftrightarrow3x\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}}\)

Vậy x = 0; x = 4; x = - 4 là nghiệm của pt.

3, \(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}}\)

Vậy x = 1; x = 1/5 là nghiệm của pt.

4, \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}}\)

Vậy x = - 5; x = 2 là nghiệm của pt.

5, \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

Vậy x = - 2 là nghiệm của pt.

\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{7^x.\left(7^2+7+1\right)}{57}=7^x\)

\(\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^3}{131}=\frac{5^{2x}\left(1+5+5^3\right)}{131}=\frac{25^x.131}{131}=25^x\)

\(\Rightarrow7^x=25^x\Rightarrow x=0\)

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