tìm x : 2x-1 +5.2x-2=\(\frac{7}{32}\)
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\(1,x^3-3x^2=0\)
\(x^2\left(x-3\right)=0\)
\(\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=3\left(TM\right)\end{cases}}}\)
\(2,3x^3-48x=0\)
\(3x\left(x^2-16\right)=0\)
\(\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x^2=16\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=\pm4\left(TM\right)\end{cases}}}}\)
\(3,5x\left(x-1\right)=x-1\)
\(5x^2-5x=x-1\)
\(5x^2-6x+1=0\)
\(5x^2-5x-x+1=0\)
\(5x\left(x-1\right)-\left(x-1\right)=0\)
\(\left(5x-1\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}5x-1=0\\x-1=0\end{cases}\orbr{\begin{cases}x=\frac{1}{5}\left(TM\right)\\x=1\left(TM\right)\end{cases}}}\)
\(4,2\left(x+5\right)-x^2-5x=0\)
\(2x+10-x^2-5x=0\)
\(-x^2-3x+10=0\)
\(-x^2-5x+2x+10=0\)
\(-x\left(x+5\right)+2\left(x+5\right)=0\)
\(\left(x+5\right)\left(2-x\right)=0\)
\(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\orbr{\begin{cases}x=-5\left(TM\right)\\x=2\left(TM\right)\end{cases}}}\)
\(5,2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x-26=0\)
\(-13\left(x+2\right)=0\)
\(x=-2\left(TM\right)\)
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.
1) \(\left(\frac{2x}{3}-3\right):\left(-10\right)=\frac{2}{5}\)
\(\Leftrightarrow-\frac{\frac{2x}{3}-3}{10}=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{\frac{2x}{3}}{10}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{2x}{3\times10}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{2x}{30}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\frac{x}{15}+\frac{3}{10}=\frac{2}{5}\)
\(\Leftrightarrow\frac{3}{10}-\frac{x}{15}=\frac{2}{5}\)
\(\Leftrightarrow-\frac{x}{15}=\frac{2}{5}-\frac{3}{10}\)
\(\Leftrightarrow-\frac{x}{15}=\frac{1}{10}\)
\(\Leftrightarrow-x=\frac{15}{10}\)
\(\Leftrightarrow-x=\frac{3}{2}\)
\(\Leftrightarrow x=-\frac{3}{2}\)
Vậy \(x=-\frac{3}{2}\)
2) \(\left|2x-1\right|+1=4\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
Vậy \(x\in\left\{2;-1\right\}\)
\(\frac{32-x}{7}=\frac{x-42}{9}\)
=\(\frac{\left(32-x\right)9}{63}=\frac{\left(x-42\right)7}{63}\)
\(\Rightarrow\)\(\left(32-x\right)9=\left(x-42\right)7\)
=\(288-x9=x7-294\)
=\(288+294=x9+x7\)
=\(x=-36\frac{6}{16}\)
=\(x\times16=-582\)
\(x=-582\div16\)
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
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/\(\left(4,5-2x\right).1\frac{4}{7}=\frac{11}{14}\)
\(\left(\frac{45}{10}-2x\right).\frac{11}{7}=\frac{11}{14}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}:\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}.\frac{7}{11}\)
\(\left(\frac{9}{2}-2x\right)=\frac{1}{2}.\frac{1}{1}=\frac{1}{2}\)
\(2x=\frac{9}{2}-\frac{1}{2}\)
\(2x=\frac{8}{2}\)
\(x=\frac{8}{2}:\frac{2}{1}=\frac{8}{2}.\frac{1}{2}\)
\(x=\frac{4}{2}.\frac{1}{1}=\frac{4}{2}=2\)
\(\left(2.8x-32\right):\frac{2}{3}=90\)
\(2.8\cdot x-32=90\cdot\frac{2}{3}\)
\(\frac{14}{5}x-32=60\)
\(\frac{14}{5}x=60+32\)
\(\frac{14}{5}x=92\)
\(x=\frac{230}{7}\)
B , c , d tương tự
2x-1+5.2x-2=2.2x-2+5.2x-2=2x-2.(5+2)=2x-2.7=7/32
=>2x-2=7/32:7=1/32
=>x-2=-5
=>x=-3
=> x =3