tìm x:
a, \(4x^2+1=8-2\sqrt{6}\)
b, \(x^2+1=6-2\sqrt{6}\)
K
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TG
16 tháng 5 2021
a) ĐK: x ≥ 2
\(\sqrt{3x-6}=3\)
\(\Leftrightarrow3x-6=9\)
<=> 3x = 15
<=> x = 5
Vậy:....
b) ĐK: 5x - 16 ≥ 0
<=> 5x ≥ 16
<=> x ≥ 16/5
\(\sqrt{5x-16}=2\)
<=> 5x - 16 = 4
<=> 5x = 20
<=> x = 4
c) ĐK: \(x^2-4x+3\ne0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
16 tháng 5 2021
bình phương hai vế ta được:
a)điều kiện của x:x≥2
3x-6=9 <=> x=5(nhận)
b)ĐK: x≥16/5
5x-16=4 <=>x=4(nhận)
c) ta có: \(\dfrac{2x-3}{\left(x-2\right)^2-1}\)= \(\dfrac{2x-3}{\left(x-3\right)\left(x-1\right)}\)
ĐKXĐ: x≠3 ;x≠1
24 tháng 11 2021
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
31 tháng 8 2021
c: Ta có: \(\sqrt{x^2}=x\)
\(\Leftrightarrow\left|x\right|=x\)
hay \(x\ge0\)
d: Ta có: \(\sqrt{x^2-2x+1}=x-1\)
\(\Leftrightarrow\left|x-1\right|=x-1\)
hay \(x\ge1\)
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)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
17 tháng 12 2021
Bài 1:
ĐKXĐ: $3-2x\geq 0\Leftrightarrow x\leq \frac{3}{2}$
Bài 2:
a. ĐKXĐ: $x\geq \frac{1}{3}$
PT $\Leftrightarrow 3x-1=2^2=4$
$\Leftrightarrow x=\frac{5}{3}$ (tm)
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{x-2}+2\sqrt{x-2}=6$
$\Leftrightarrow 3\sqrt{x-2}=6$
$\Leftrightarrow \sqrt{x-2}=2$
$\Leftrightarrow x-2=4$
$\Leftrightarrow x=6$ (tm)
PA
25 tháng 7 2017
\(x^2-4x-6=\sqrt{2x^2-8x+12}\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(6x+6+\sqrt{2x^2-8x+12}\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-\dfrac{36x^2+72x+36-\left(2x^2-8x+12\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}=0\)
\(\Leftrightarrow x\left(x+2\right)-\dfrac{2\left(17x+6\right)\left(x+2\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}=0\)
\(\Leftrightarrow\left(x+2\right)\left[x-\dfrac{2\left(17x+6\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}\right]=0\)
Pt \(x-\dfrac{2\left(17x+6\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}\) vô nghiệm
=> x + 2 = 0
<=> x = - 2 (nhận)
PA
25 tháng 7 2017
\(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-2}-2\right|+\left|\sqrt{x-2}-3\right|=1\)
Ta có:
\(VT=\left|\sqrt{x-2}-2\right|+\left|3-\sqrt{x-2}\right|\ge\left|\sqrt{x-2}-2+3-\sqrt{x-2}\right|=1\)
Dấu "=" xảy ra khi \(\left(\sqrt{x-2}-2\right)\left(3-\sqrt{x-2}\right)\ge0\)
Bảng xét dấu:
Vậy \(6\le x\le11\)
\(a,4x^2+1=8-2\sqrt{6}.\)
\(\Leftrightarrow4x^2=7-2\sqrt{6}\)
\(\Leftrightarrow x^2=\frac{7-2\sqrt{6}}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{7-2\sqrt{6}}}{2}\\x=\frac{-\sqrt{7-2\sqrt{6}}}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{\left(\sqrt{6}-1\right)^2}}{2}\\x=\frac{-\sqrt{\left(\sqrt{6}-1\right)^2}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{6}-1}{2}\\x=\frac{1-\sqrt{6}}{2}\end{cases}}}\)
\(b,x^2+1=6-2\sqrt{6}.\)
\(\Leftrightarrow x^2=5-2\sqrt{6}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5-2\sqrt{6}}\\x=-\sqrt{5-2\sqrt{6}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\x=-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{3}-\sqrt{2}\\x=\sqrt{2}-\sqrt{3}\end{cases}}\)
Vậy ...............