\(\sqrt{\text{4x-20}}+\sqrt{x-5}=6\)
Câu 2: Tìm x biết:
a. \(\sqrt{x-1}=2\)
b. \(\sqrt{3x+1}=\sqrt{4x-3}\)
c. \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
d. \(\sqrt{x^2-4x+4}=\sqrt{6+2\sqrt{5}}\)
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
d) \(x-5\sqrt{x}+6=0\)
e) \(\sqrt{x-1}+\dfrac{3}{2}\sqrt{4x-4}-\dfrac{2}{5}\sqrt{25x-25}=4\)
f) \(\sqrt{x-5}+\sqrt{4x-20}-\dfrac{1}{3}\sqrt{9x-45}=6\)
\(d,ĐK:x\ge0\\ PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+\dfrac{3}{2}\cdot2\sqrt{x-1}-\dfrac{2}{5}\cdot5\sqrt{x-1}=4\\ \Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\\ \Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\\ f,ĐK:x\ge5\\ PT\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\\ \Leftrightarrow2\sqrt{x-5}=6\Leftrightarrow\sqrt{x-5}=3\\ \Leftrightarrow x-5=9\Leftrightarrow x=14\left(tm\right)\)
2.tìm x
a)\(\sqrt{x^2-6x+9}\)
b)\(\sqrt{x^2-2x+1}\)
c)\(\sqrt{4x+12}-3\sqrt{x+3}+7\sqrt{9x+27}=20\)
d)\(\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=6\)
a) \(\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x^2-2.x.3+3^2\right)}\)
\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x
⇒x∈\(R\)
b) \(\sqrt{x^2-2x+1}\)
\(=\sqrt{\left(x^2-2.x.1+1^2\right)}\)
\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x
⇒x∈\(R\)
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\) tìm x
Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
hay x=-1
\(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\cdot\sqrt{9x+45}=6\left(x\ge-5\right)\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\cdot\sqrt{x+5}=6\)
\(\Leftrightarrow3\cdot\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=2^2=4\)
\(\Leftrightarrow x=-1\left(N\right)\)
Tìm x biết:
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+15}=6\)
ĐKXĐ:\(x\ge-5\)
`sqrt{4x+20}-3sqrt{5+x}+4/3sqrt{9x+15}=6(x>=-5)`
`<=>sqrt{4(x+5)}-3sqrt{x+5}+4/3sqrt{9(x+5)}=6`
`<=>2sqrt{x+5}-3sqrt{x+5}+4sqrt{x+5}=6`
`<=>3sqrt{x+5}=6`
`<=>sqrt{x+5}=2`
`<=>x+5=4`
`<=>x=-1(tm)`
Vậy `x=-1`
Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\cdot\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow0\cdot\sqrt{x+5}=6\left(vôlý\right)\)
a \(\sqrt{4x-20}+\sqrt{x-5}=4+3\sqrt{\dfrac{x-5}{9}}\)
b \(\sqrt{4x-20}+\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{4x-45}=4\)
Lời giải:
a. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}=4+3.\sqrt{\frac{1}{9}}.\sqrt{x-5}$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}=4+\sqrt{x-5}$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x-5=4$
$\Leftrightarrow x=9$ (tm)
b. Sửa đoạn 4x-45 thành 4x-20.
ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{4}.\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}+\frac{1}{3}\sqrt{x-5}-\frac{2}{3}\sqrt{x-5}=4$
$\Leftrightarrow \frac{5}{3}\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=\frac{12}{5}$
$\Leftrightarrow x-5=\frac{144}{25}=5,76$
$\Leftrightarrow x=10,76$ (tm)
a)\(\sqrt{4-5x}=12\) tìm x
b)\(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)
c)\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
a) Ta có: \(\sqrt{4-5x}=12\)
\(\Leftrightarrow4-5x=144\)
\(\Leftrightarrow5x=-140\)
hay x=-28
b) Ta có: \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}+10=10+4\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}=4\sqrt{6}\)
\(\Leftrightarrow3x=96\)
hay x=32
c) Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
a)
\(\sqrt{\left(5x+2\right)^2}\)=3
b)
\(\sqrt{4x+20}\)+3\(\sqrt{5+x}\)-4=6
Lời giải:
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow |5x+2|=3$
$\Leftrightarrow 5x+2=\pm 3$
$\Leftrightarrow x=-1$ hoặc $x=\frac{1}{5}$ (tm)
b. ĐKXĐ: $x\geq -5$
PT $\Leftrightarrow \sqrt{4(x+5)}+3\sqrt{5+x}=10$
$\Leftrightarrow 2\sqrt{x+5}+3\sqrt{x+5}=10$
$\Leftrightarrow 5\sqrt{x+5}=10$
$\Leftrightarrow \sqrt{x+5}=2$
$\Leftrightarrow x+5=4$
$\Leftrightarrow x=-1$ (tm)
\(A=\sqrt{4x+20}-2\sqrt{x+5}+\sqrt{9x+45}\)
a) Rút gọn A
b) Tìm x để A=6
\(a,ĐK:x\ge-5\\ A=2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=3\sqrt{x+5}\\ b,A=6\Leftrightarrow\sqrt{x+5}=\dfrac{6}{3}=2\\ \Leftrightarrow x+5=4\\ \Leftrightarrow x=-1\left(tm\right)\)
\(A=2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}\)
\(A=3\sqrt{x+5}\)
\(3\sqrt{x+5}=6\)
\(\sqrt{x+5}=2\)
\(\left\{{}\begin{matrix}2\ge0\left(ld\right)\\x+5=4\end{matrix}\right.\)
\(x=-1\)