giải phương trình
a, (x-2)^2-(x-3)(x+3)=6
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NX
1
1 tháng 2 2021
a)
ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(\dfrac{2x}{x-3}=\dfrac{x^2+11x-6}{x^2-9}\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+11x-6}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(2x^2+6x=x^2+11x-6\)
\(\Leftrightarrow2x^2+6x-x^2-11x+6=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
Vậy: S={2}
b) Ta có: \(3x^2+\left(1-\sqrt{3}\right)x+\sqrt{3}-4=0\)
\(\Leftrightarrow3x^2-\left(\sqrt{3}-1\right)x+\sqrt{3}-4=0\)
\(\Leftrightarrow3x^2-\left(\sqrt{3}-1\right)x+\sqrt{3}-1-3=0\)
\(\Leftrightarrow\left(3x^2-3\right)-\left(\sqrt{3}-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-\left(\sqrt{3}-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3-\sqrt{3}+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+4-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+4-\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=\sqrt{3}-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{\sqrt{3}-4}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{1;\dfrac{\sqrt{3}-4}{3}\right\}\)
10 tháng 1 2022
a: \(\Leftrightarrow\dfrac{15-2x-1}{5}>\dfrac{x+3}{4}\)
\(\Leftrightarrow\dfrac{-8x+56}{20}>\dfrac{5x+15}{20}\)
=>-8x+56>5x+15
=>-11x>-41
hay x<41/11
b: \(\Leftrightarrow\dfrac{5x+5-6}{6}< \dfrac{4x+4}{6}\)
=>5x-1<4x+4
=>x<5
10 tháng 1 2022
\(3-\dfrac{2x+1}{5}>x+\dfrac{3}{4}.\)
\(\Leftrightarrow\dfrac{14-2x}{5}-x-\dfrac{3}{4}>0.\)
\(\Leftrightarrow\dfrac{56-8x-20x-15}{20}>0.\)
\(\Rightarrow-28x+41>0.\)
\(\Leftrightarrow-28x>-41.\)
\(\Leftrightarrow x< \dfrac{41}{28}.\)
AT
2
H
7 tháng 4 2021
a) 5(x-1)(x+1)=5x^2+3x-2
<=> (5x-5)(x+1) = (x+1)(5x-2)
<=> (x+1)(5x-5) - (x+1)(5x-2)=0
<=> (x+1)(5x-5-5x+2)=0
<=> (x+1).(-3)=0
<=> x+1=0<=> x=-1
BC
2
LT
6 tháng 3 2021
a) \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow x^2-4x+4-x^2+16=0\)
\(\Leftrightarrow20-4x=0\)
\(\Leftrightarrow4x=20\)
\(\Leftrightarrow x=5\)
Vậy S = {5}
b) ĐKXĐ: \(x\ne0;x\ne-2\)
\(\dfrac{x+2}{x}=\dfrac{\left(x+1\right)\left(x+4\right)}{x^2+2x}+\dfrac{x}{x+2}\)
\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{x^2+4x+x+4+x^2}{x\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{2x^2+5x+4}{x\left(x+2\right)}\)
\(\Rightarrow x\left(x+2\right)^2=x\left(2x^2+5x+4\right)\)
\(\Leftrightarrow x^3+4x^2+4x=2x^3+5x^2+4x\)
\(\Leftrightarrow x^3+x^2=0\)
\(\Leftrightarrow x^2\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(TM\right)\end{matrix}\right.\)
Vậy S = {-1}
c) Câu này mình không chắc về đề lắm! Bạn dùng ô chữ M bị ngược để viết lại đề nhé!
6 tháng 3 2021
a) Ta có: \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow x^2-4x+4=x^2-16\)
\(\Leftrightarrow x^2-4x+4-x^2+16=0\)
\(\Leftrightarrow-4x+20=0\)
\(\Leftrightarrow-4x=-20\)
hay x=5
Vậy: S={5}
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.
4 tháng 7 2021
a) Ta có: \(\sqrt{\left(x-3\right)^2}=2\)
\(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) ĐKXĐ: \(x\ge-2\)
Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+\dfrac{4}{5}\cdot5\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7(thỏa ĐK)
TG
4 tháng 7 2021
a) \(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy:.....
b) ĐKXĐ: x ≥ -2
\(\Leftrightarrow\sqrt{9}.\sqrt{x+2}-5.\sqrt{x+2}+\dfrac{4}{5}.\sqrt{25}.\sqrt{x+2}=6\)
<=> \(\sqrt{x+2}.\left(3-5+\dfrac{4}{5}.5\right)=6\)
\(\Leftrightarrow2.\sqrt{x+2}=6\)
\(\Leftrightarrow\sqrt{x+2}=3\)
<=> x + 2 = 9
<=> x = 7
8T
1
18 tháng 1 2022
\(\text{2x - (x - 3)(5 - x) = (x+4)}^2.\)
\(\Leftrightarrow2x-\left(5x-x^2-15+3x\right)=x^2+8x+16.\)
\(\Leftrightarrow2x-5x+x^2+15-3x-x^2-8x-16=0.\)
\(\Leftrightarrow-14x-1=0.\Leftrightarrow x=\dfrac{-1}{14}.\)
\(\text{(4x + 1)(x - 2) + 25 = (2x+3)}^2-4x.\)
\(\Leftrightarrow4x^2-8x+x-2+25=4x^2+12x+9-4x.\)
\(\Leftrightarrow-15x+14=0.\Leftrightarrow x=\dfrac{14}{15}.\)
13 tháng 6 2023
b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1
=>-3/4x+1/6x=2+5-4/5=24/5
=>x=-288/35
c: =>6x^2+3x-30x-15=6x^2+10x-21x-35
=>-27x-15=-11x-35
=>-16x=-20
=>x=5/4
6 tháng 2 2022
a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
=>x-2=16
hay x=18
b: \(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)
c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
\(\Leftrightarrow4\sqrt{x-2}=40\)
=>x-2=100
hay x=102
d: =>5x-6=9
hay x=3
MH
6 tháng 2 2022
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\Leftrightarrow x^2-2x+1-x^2+9=-2x+10=6\Leftrightarrow-2x=-4\Leftrightarrow x=2\)