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\(\sqrt{4x^2-4x+1}=\sqrt{x^2+10x+25}\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x+5\right)^2}\)
\(\Leftrightarrow\left|2x-1\right|=\left|x+5\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-\left(x+5\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-x-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)
a)
\(\sqrt{x+3}+2\sqrt{4\left(x+3\right)}-\frac{1}{3}\sqrt{9\left(x+3\right)}=8\)
\(\sqrt{x+3}+2\cdot2\sqrt{x+3}-\frac{1}{3}\cdot3\sqrt{x+3}=8\)
\(\sqrt{x+3}+4\sqrt{x+3}-\sqrt{x+3}=8\)
\(4\sqrt{x+3}=8\)
\(\sqrt{x+3}=2\)
\(\orbr{\begin{cases}2\ge0\left(llđ\right)\\x+3=2^2\end{cases}}\)
\(x+3=4\)
\(x=1\)
b)
\(\orbr{\begin{cases}x^2+10x+25\ge0\\4x^2-4x+1=x^2+10x+25\end{cases}}\)
\(\orbr{\begin{cases}\left(x+5\right)^2\ge0\left(lld\right)\\3x^2-6x-24=0\end{cases}}\)
\(\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)
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a) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)
ĐK : x ≥ 0
⇔ \(\frac{3}{4}\sqrt{x}-\sqrt{3^2x}-\frac{1}{4}\sqrt{3^2x}=-5\)
⇔ \(\frac{3}{4}\sqrt{x}-3\sqrt{x}-\frac{1}{4}\cdot3\sqrt{x}=-5\)
⇔ \(-\frac{9}{4}\sqrt{x}-\frac{3}{4}\sqrt{x}=-5\)
⇔ \(-3\sqrt{x}=-5\)
⇔ \(\sqrt{x}=15\)
⇔ \(x=225\)( tm )
b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
ĐK : x ≤ 3
⇔ \(\sqrt{3-x}-\sqrt{3^2\left(3-x\right)}+\frac{5}{4}\sqrt{4^2\left(3-x\right)}=6\)
⇔ \(\sqrt{3-x}-3\sqrt{3-x}+\frac{5}{4}\cdot4\sqrt{3-x}=6\)
⇔ \(-2\sqrt{3-x}+5\sqrt{3-x}=6\)
⇔ \(3\sqrt{3-x}=6\)
⇔ \(\sqrt{3-x}=2\)
⇔ \(3-x=4\)
⇔ \(x=-1\)( tm )
c) \(\sqrt{9x^2+12x+4}=4\)
⇔ \(\sqrt{\left(3x+2\right)^2}=4\)
⇔ \(\left|3x+2\right|=4\)
⇔ \(\orbr{\begin{cases}3x+2=4\\3x+2=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-2\end{cases}}\)
d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)
ĐK : x ≥ 1
⇔ \(\frac{1}{3}\sqrt{x-1}+2\sqrt{2^2\left(x-1\right)}-12\sqrt{\left(\frac{1}{5}\right)^2\cdot\left(x-1\right)}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+2\cdot2\sqrt{x-1}-12\cdot\frac{1}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{29}{15}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\sqrt{x-1}=1\)
⇔ \(x-1=1\)
⇔ \(x=2\)( tm )
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Bài 2 giải như sau (sau khi tác giả đã sửa): Điều kiện \(x,y>0.\)
Từ hệ ta suy ra \(1+\frac{3}{x+3y}=\frac{2}{\sqrt{x}},1-\frac{3}{x+3y}=\frac{4\sqrt{2}}{\sqrt{7y}}.\) Cộng và trừ hai phương trình, chia cả hai vế cho 2, ta sẽ được 2 phương trình \(1=\frac{1}{\sqrt{x}}+\frac{2\sqrt{2}}{\sqrt{7y}},\frac{3}{x+3y}=\frac{1}{\sqrt{x}}-\frac{2\sqrt{2}}{\sqrt{7y}}.\) Nhân hai phương trình với nhau, vế theo vế, ta được
\(\frac{3}{x+3y}=\frac{1}{x}-\frac{8}{7y}\to21xy=\left(x+3y\right)\left(7y-8x\right)\to21y^2-38xy-8x^2=0\to x=\frac{y}{2},x=-\frac{21}{4}y.\)
Đến đây ta được y=2x (trường hợp kia loại). Từ đó thế vào ta được \(1+\frac{3}{7x}=\frac{2}{\sqrt{x}}\to7x-14\sqrt{x}+3=0\to\sqrt{x}=\frac{7\pm2\sqrt{7}}{2}\to...\)
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\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
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\(A=\left(\frac{\sqrt{3}}{x^2+x\sqrt{x}+3}+\frac{3}{x^3-\sqrt{27}}\right)\left(\frac{x}{\sqrt{3}}+\frac{\sqrt{3}}{x}+1\right)\)
\(\Leftrightarrow A=\left[\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}+\frac{3}{\left(x-\sqrt{3}\right)\left(x+x\sqrt{3}+3\right)}\right]\left(\frac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\right)\)
\(\Leftrightarrow A=\frac{x\sqrt{3}-3+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}.\frac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(\Leftrightarrow A=\frac{1}{x-\sqrt{3}}\)
ĐKXĐ : \(x\ge3\)
\(\sqrt{9x-27}-2\sqrt{\frac{x-3}{4}}=1\)
\(\Leftrightarrow\)\(3\sqrt{x-3}-\sqrt{x-3}=1\)
\(\Leftrightarrow\)\(2\sqrt{x-3}=1\)
\(\Leftrightarrow\)\(\sqrt{x-3}=\frac{1}{2}\)
\(\Leftrightarrow\)\(x=\frac{13}{4}\) ( tm đkxđ )
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