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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 $x=3$
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.
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)
\(a,\left(3x+1\right)^2-\left(2x-5\right)^2=0\\ \Leftrightarrow\left(3x+1+2x-5\right)\left(3x+1-2x+5\right)=0\\ \Leftrightarrow\left(5x-4\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-6\end{matrix}\right.\\ b,\left(x+3\right)\left(4-3x\right)=x^2+6x+9\\ \Leftrightarrow\left(x+3\right)\left(4-3x\right)-\left(x+3\right)^2=0\\ \Leftrightarrow\left(x+3\right)\left(4-3x-x-3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(1-4x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{4}\end{matrix}\right.\)
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
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
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 \(S=\left\{\dfrac{1}{2};-8;5\right\}\)
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 \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)
câu này xài cách đặt ẩn giống câu trên luôn
b) Đặt n = x2-3x+3 ta được
n(n+x)=2x2
n2 +nx-2x2=0
n^2-1nx+2nx-2x^2=0
n(n-x)+2x(n-x)=0
(n+2x)(n-x)=0
(x^2-3x+3+2x)(x^2-3x+3-x)=0
(x^2-x+3)(x^2-4x+3)=0
mà x^2-x+3 =0
x^2-1/2.2x+1/4-1/4+3=0
(x+1/2)^2+11/4 >0( loại)
Vậy ta còn
x^2-4x+3=0
x^2-1x-3x+3=0
(x-1)(x-3)=0
<=> x-1=0 hay x-3=0
x=1 hay x=3
Vậy S= (1;3)
a) (x -1)(x-6)(x-5)(x-2)=252
<=>( x^2-7x+6)(x^2-7x+10)=252
Đặt n=x^2-7x+6 ta được :
n(n+4)=252
n^2+4n-252=0
n^2-14n+18n-252=0
n(n-14)+18(n-14)=0
(n+18)(n-14)=0
r tới đây bạn tự giải tiếp nha, mình đánh máy ko quen nên hơi lâu, với bạn tự thêm dấu tương đương nữa, chờ mình câu2