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a) \(\left(x+2\right)^2-\left(3x-7\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=3x-7\\x+2=-3x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3x=-2-7\\x+3x=-2+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=-9\\4x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Mấy câu kia tương tự.
a) \(\left(x+2\right)^2-\left(3x-7\right)^2=0\)
\(\Leftrightarrow\left(x+2-3x+7\right)\left(x+2+3x-7\right)=0\)
\(\Leftrightarrow\left(-2x+9\right)\left(4x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x+9=0\\4x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=-9\\4x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9}{-2}=\dfrac{9}{2}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Vậy \(x=\dfrac{9}{2}\) hoặc \(x=\dfrac{5}{4}\)
b) lộn đề à
c) \(25\left(x-3\right)^2-49\left(2x+1\right)^2=0\)
\(\Leftrightarrow5^2\left(x-3\right)^2-7^2\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[5\left(x-3\right)\right]^2-\left[7\left(2x+1\right)\right]^2=0\)
\(\Leftrightarrow\left(5x-15\right)^2-\left(14x+7\right)^2=0\)
\(\Leftrightarrow\left(5x-15-14x-7\right)\left(5x-15+14x+7\right)=0\)
\(\Leftrightarrow\left(-9x-22\right)\left(19x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-9x-22=0\\19x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-9x=22\\19x=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{22}{-9}=\dfrac{-22}{9}\\x=\dfrac{8}{19}\end{matrix}\right.\)
Vậy \(x=\dfrac{-22}{9}\) hoặc \(x=\dfrac{8}{19}\)
d) \(9\left(3x-2\right)^2=121\left(1-4x\right)^2\)
\(\Leftrightarrow9\left(3x-2\right)^2-121\left(1-4x\right)^2=0\)
\(\Leftrightarrow3^2\left(3x-2\right)^2-11^2\left(1-4x\right)^2=0\)
\(\Leftrightarrow\left[3\left(3x-2\right)\right]^2-\left[11\left(1-4x\right)\right]^2=0\)
\(\Leftrightarrow\left(9x-6\right)^2-\left(11-44x\right)^2=0\)
\(\Leftrightarrow\left(9x-6-11+44x\right)\left(9x-6+11-44x\right)=0\)
\(\Leftrightarrow\left(53x-17\right)\left(-35x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}53x-17=0\\-35x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}53x=17\\-35x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{53}\\x=\dfrac{-5}{-35}=\dfrac{1}{7}\end{matrix}\right.\)
Vậy \(x=\dfrac{17}{53}\) hoặc \(x=\dfrac{1}{7}\)

\(a,3.\left(2x+8\right)-\left(5x+2\right)=0\)
\(\Rightarrow6x+24-5x+2=0\)
\(\Rightarrow6x-5x=0-24-2\)
\(\Rightarrow x=-26\)
\(5.\left(7-3x\right)+7.\left(2+2x\right)=0\)
\(\Rightarrow35-15x+14+14x=0\)
\(\Rightarrow-15x+14x=0-35-14\)
\(\Rightarrow-x=-49\)
\(\Rightarrow x=49\)
Tìm x
a, 3.(2x + 8) - (5x + 2)=0 ............. Hok Tốt nhé ..............
\(\Leftrightarrow3x+24-5x-2=0\) ........ Nhớ k cho mik nhé .........
\(\Leftrightarrow x+22=0\)
\(\Rightarrow x=-22\)
b, 5.(7 - 3x) + 7.( 2 + 2x)
\(\Leftrightarrow35-15x+14+14x\)
\(\Leftrightarrow-15x+14x=0-35-14\)
\(\Leftrightarrow-x=-49\)
\(\Rightarrow x=49\)

a) <=> 2x - 1 = 0 hoặc 5x + 2 = 0
<=> 2x = 1 hoặc 5x = -2
<=> x = \(\frac{1}{2}\) hoặc x = \(-\frac{2}{5}\)
b) <=> 3/7 . (1 + 1/x) = 1/4
=> 1 + 1/x = 7/12 <=> 1/x = -5/12
<=> -5/-5x = -5/12 <=> -5x = 12
<=> x = \(-\frac{12}{5}\)
c) Dễ thấy 3x + 5 > 2x - 3
Để (3x + 5)( 2x - 3) < 0 thì 3x + 5 > 0 và 2x - 3 < 0
<=> x > -5/3 và x < 3/2
Vậy \(-\frac{5}{3}< x< \frac{3}{2}\)
a) (2x-1).(5x+2) = 0
\(\Leftrightarrow\) 2x-1 = 0 hoặc 5x+2 = 0
\(\Leftrightarrow\) 2x = 1 hoặc 5x = -2
\(\Leftrightarrow\) x = \(\frac{1}{2}\) hoặc x = \(\frac{-2}{5}\)
b) \(\frac{3}{7}+\frac{3}{7}:x=\frac{-1}{2}-\left(\frac{-3}{4}\right)\)
\(\frac{3}{7}+\frac{3}{7}:x=\frac{-1}{2}+\frac{3}{4}\)
\(\frac{3}{7}+\frac{3}{7}:x=\frac{1}{4}\)
\(\frac{3}{7}:x=\frac{1}{4}-\frac{3}{7}\)
\(\frac{3}{7}:x=\frac{-5}{28}\)
\(x=\frac{3}{7}:\frac{-5}{28}\)
\(x=\frac{-12}{5}\)
`(3x+2)(-2/5x-7)=0`
\(\Rightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=-2\\-\dfrac{2}{5}x=7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-\dfrac{35}{2}\end{matrix}\right.\)