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19 tháng 2 2022
b: \(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
=>-6x+16=0
=>-6x=-16
hay x=8/3(nhận)
c: \(\Leftrightarrow\dfrac{x+1+x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+2}\)
\(\Leftrightarrow2x\left(x+2\right)=2\left(x^2-1\right)\)
\(\Leftrightarrow2x^2+4x-2x^2+2=0\)
=>4x+2=0
hay x=-1/2(nhận)
AH
Akai Haruma
Giáo viên
27 tháng 4 2023
Bài 1:
a.
$(4x^2+4x+1)-x^2=0$
$\Leftrightarrow (2x+1)^2-x^2=0$
$\Leftrightarrow (2x+1-x)(2x+1+x)=0$
$\Leftrightarrow (x+1)(3x+1)=0$
$\Rightarrow x+1=0$ hoặc $3x+1=0$
$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$
b.
$x^2-2x+1=4$
$\Leftrightarrow (x-1)^2=2^2$
$\Leftrightarrow (x-1)^2-2^2=0$
$\Leftrightarrow (x-1-2)(x-1+2)=0$
$\Leftrightarrow (x-3)(x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-1$
c.
$x^2-5x+6=0$
$\Leftrightarrow (x^2-2x)-(3x-6)=0$
$\Leftrightarrow x(x-2)-3(x-2)=0$
$\Leftrightarrow (x-2)(x-3)=0$
$\Leftrightarrow x-2=0$ hoặc $x-3=0$
$\Leftrightarrow x=2$ hoặc $x=3$
AH
Akai Haruma
Giáo viên
27 tháng 4 2023
2c.
ĐKXĐ: $x\neq 0$
PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$
$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$
$\Leftrightarrow x=-4$ (tm)
2d.
ĐKXĐ: $x\neq 2$
PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$
$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$
$\Rightarrow 3x-5=3-x$
$\Leftrightarrow 4x=8$
$\Leftrightarrow x=2$ (không tm)
Vậy pt vô nghiệm.
LV
1
26 tháng 2 2022
a: =>5x-5+17x=1-12x-4
=>22x-5=-12x-3
=>34x=2
hay x=1/17
b: =>\(\left(x-3\right)^2-4x\left(x-3\right)=0\)
=>(x-3)(-3x-3)=0
=>x=3 hoặc x=-1
c: =>(x-4)(x-6)=0
=>x=4 hoặc x=6
5 tháng 1 2021
a) ĐKXĐ: x≠-5
Ta có: \(\dfrac{2x-5}{x+5}=4\)
\(\Leftrightarrow2x-5=4\left(x+5\right)\)
\(\Leftrightarrow2x-5=4x+20\)
\(\Leftrightarrow2x-5-4x-20=0\)
\(\Leftrightarrow-2x-25=0\)
\(\Leftrightarrow-2x=25\)
hay \(x=\dfrac{-25}{2}\)(nhận)
Vậy: \(S=\left\{-\dfrac{25}{2}\right\}\)
b) ĐKXĐ: x≠0
Ta có: \(\dfrac{x^2-4}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-4\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-8=2x^2+3x\)
\(\Leftrightarrow2x^2-8-2x^2-3x=0\)
\(\Leftrightarrow-3x-8=0\)
\(\Leftrightarrow-3x=8\)
hay \(x=\dfrac{-8}{3}\)(nhận)
Vậy: \(S=\left\{-\dfrac{8}{3}\right\}\)
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-5\right\}\)
Ta có: \(\dfrac{2x+3}{2x-1}=\dfrac{x-3}{x+5}\)
\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)
\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)
\(\Leftrightarrow2x^2+13x+15-2x^2+7x-3=0\)
\(\Leftrightarrow20x+12=0\)
\(\Leftrightarrow20x=-12\)
hay \(x=-\dfrac{3}{5}\)(nhận)
Vậy: \(S=\left\{-\dfrac{3}{5}\right\}\)
d) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(x+7\right)\left(6x+1\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+x+42x+7\)
\(\Leftrightarrow6x^2-13x+6=6x^2+43x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
NT
1
14 tháng 3 2021
Ta có (\(^{x^{2^{ }}^{ }+3x}\)) (\(^{x^{2^{ }}+3x+4}\))
Đặt \(x^{2^{ }^{ }}+3x\) là a ta có
a.(a+4)=-4
4a+\(a^2\) -4=0
\(^{ }\left(a-2\right)^2\)=0
Suy ra a=2
hay \(x^{2^{ }^{ }^{ }}+3x=2\)
\(x^2+3x-2=0\)
𝑥=−3±17√/2
23 tháng 2 2022
a: \(\Leftrightarrow3x+2\left(x+2\right)=5\left(x-1\right)\)
=>3x+2x+4=5x-5
=>4=-5(vô lý)
b: \(\Leftrightarrow\dfrac{2}{x\left(x+4\right)}-\dfrac{3x}{x+4}=-3\)
\(\Leftrightarrow2-3x^2=-3x\left(x+4\right)\)
\(\Leftrightarrow2-3x^2+3x^2+12x=0\)
=>12x+2=0
hay x=-1/6
TG
14 tháng 1 2021
a) (x - 7)(2x + 8) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\2x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy: S = {7; -4}
b) Tương tự câu a
c) (x - 1)(2x + 7)(x2 + 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\\x^2+2=0\end{matrix}\right.\)
Mà: x2 + 2 > 0 với mọi x
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{2}\right\}\)
d) (2x - 1)(x + 8)(x - 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-8\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};-8;5\right\}\)
TT
14 tháng 1 2021
a/ Pt \(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy \(S=\left\{7;-4\right\}\)
b/ pt \(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\5x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)
c/ pt \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\) (\(x^2+2>0\forall x\))\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
d/ pt \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)
\(a,\left(3x+1\right)^2-\left(2x-5\right)^2=0\\ \Leftrightarrow\left(3x+1+2x-5\right)\left(3x+1-2x+5\right)=0\\ \Leftrightarrow\left(5x-4\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-6\end{matrix}\right.\\ b,\left(x+3\right)\left(4-3x\right)=x^2+6x+9\\ \Leftrightarrow\left(x+3\right)\left(4-3x\right)-\left(x+3\right)^2=0\\ \Leftrightarrow\left(x+3\right)\left(4-3x-x-3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(1-4x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{4}\end{matrix}\right.\)