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f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
2:
a: =>2x^2-4x-2=x^2-x-2
=>x^2-3x=0
=>x=0(loại) hoặc x=3
b: =>(x+1)(x+4)<0
=>-4<x<-1
d: =>x^2-2x-7=-x^2+6x-4
=>2x^2-8x-3=0
=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)
c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)
a.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$
$\Leftrightarrow \sqrt{2x}=3$
$\Leftrightarrow 2x=9$
$\Leftrightarrow x=\frac{9}{2}$ (tm)
b.
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$
$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$
$\Leftrightarrow 3\sqrt{x+2}=15$
$\Leftrightarrow \sqrt{x+2}=5$
$\Leftrightarrow x+2=25$
$\Leftrightarrow x=23$ (tm)
c.
$\sqrt{(x-2)^2}=12$
$\Leftrightarrow |x-2|=12$
$\Leftrightarrow x-2=12$ hoặc $x-2=-12$
$\Leftrightarrow x=14$ hoặc $x=-10$
e.
PT $\Leftrightarrow |2x-1|-x=3$
Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$
$\Leftrightarrow x=4$ (tm)
Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
b: Ta có: \(\sqrt{x^2-6x+9}-\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{2}+1}=0\)
\(\Leftrightarrow x^2-6x+9=3\)
\(\Leftrightarrow x^2-6x+6=0\)
\(\text{Δ}=\left(-6\right)^2-4\cdot1\cdot6=36-24=12\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{3}}{2}=3-\sqrt{3}\\x_2=3+\sqrt{3}\end{matrix}\right.\)
a) ĐK: \(x\ge5\)
\(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)
\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)
\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\) (t/m)
Vậy
b) \(-5x+7\sqrt{x}=-12\)
\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
đến đây tự làm
c) d) e) bạn bình phương lên
f) \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)
\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)
\(\ge\sqrt{9}+\sqrt{25}=8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)
Vậy...
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
Đăng 1 lúc mà nhiều thế. Lần sau đăng 1 câu thôi b.
b/ \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)
Ta có: \(VT\ge1+2+\sqrt{5}=3+\sqrt{5}\)
Dấu = xảy ra khi \(x=2\)
c/ \(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x-8}=\sqrt{3-\left(x-1\right)^2}+\sqrt{1-\left(x+3\right)^2}\)
\(\le1+\sqrt{3}\)
Dấu = không xảy ra nên pt vô nghiệm
Câu d làm tương tự
\(a,\sqrt{x^2-4}-x^2+4=0\)
\(\Leftrightarrow\sqrt{x^2-4}=x^2-4\)
\(\Leftrightarrow x^2-4=\left(x-4\right)^2\)
\(\Leftrightarrow x^2-4-x^4+8x^2-16=0\)
\(\Leftrightarrow-x^4-7x^2-20=0\)
\(\Leftrightarrow-\left(x^4+7x^2+\frac{49}{4}\right)-\frac{31}{4}=0\)
\(\Leftrightarrow-\left(x^2+\frac{7}{2}\right)^2=\frac{31}{4}\)
\(\Leftrightarrow\left(x^2+\frac{7}{2}\right)=-\frac{31}{4}\)
\(\Rightarrow\)pt vô nghiệm
a) \(\sqrt{3}x-\sqrt{12}=0< =>\sqrt{3}x=\sqrt{12}=>x=2\)
Vay S = { 2 }
b) \(\sqrt{2}x+\sqrt{2}=\sqrt{8}+\sqrt{18}< =>\sqrt{2}x=\sqrt{8}+\sqrt{18}-\sqrt{2}< =>\sqrt{2}x=2\sqrt{2}+3\sqrt{2}-\sqrt{2}\) <=> \(\sqrt{2}x=4\sqrt{2}=>x=4\)
Vay S = { 4 }
c) \(\sqrt{5}x^2-\sqrt{20}=0< =>\sqrt{5}x^2=\sqrt{20}< =>x^2=2=>x=\sqrt{2}\)
Vay S = {\(\sqrt{2}\) }
d) \(\sqrt{x^2+6x+9}=3x+6< =>\sqrt{\left(x+3\right)^2}=3x+6< =>x+3=3x+6< =>-2x=\) \(3=>x=-\dfrac{3}{2}\)
Vay S = { - 3/2 }
e) \(\sqrt{x^2-4x+4}-2x+5=0< =>\sqrt{\left(x-2\right)^2}-2x+5=0< =>x-2-2x+5=0\) <=> \(-x+3=0< =>-x=-3=>x=3\)
Vay S = { 3 }
F) \(\sqrt{\dfrac{2x-3}{x-1}}=2\)
<=> \(\dfrac{2x-3}{x-1}=4< =>2x-3=4x-4< =>-2x=-1=>x=\dfrac{1}{2}\)
Vay S = { 1/2 }
g) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2< =>\sqrt{\dfrac{2x-3}{x-1}}=2< =>\dfrac{2x-3}{x-1}=4< =>2x-3=4x-4< =>-2x=-1=>x=\dfrac{1}{2}\)
bạn chưa có ĐKXĐ nên chưa xét kết quả có đúng vs Đk ko, có vài câu sai kết quả