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Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
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
Ta có: \(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}+1}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
\(\Leftrightarrow-\left(x^2-2x\right)+\sqrt{6\left(x^2-2x\right)+7}=0\) ĐK \(\sqrt{6x^2-12x+7}\ge0\)
Đặt \(t=x^2-2x\left(t\ge0\right)\Leftrightarrow pt:-t+\sqrt{6t+7}=0\Leftrightarrow\sqrt{6t+7}=t\\ 6t+7-t^2=0\Leftrightarrow\left[\begin{array}{nghiempt}t=7\left(tm\right)\\t=-1\left(ktm\right)\end{array}\right.\)
Với \(t=7\Leftrightarrow x^2-2x-7=0\Leftrightarrow x=1\pm2\sqrt{2}\left(tm\right)\)
Vậy S={\(1\pm2\sqrt{2}\)}
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
3.
ĐKXĐ: \(x\ge-1;x\ne13\)
\(\left(x+2\right)\left(\sqrt{x+1}-2\right)=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+2\right)\sqrt{x+1}-2x-4=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+1\right)\sqrt{x+1}+x+1-\left(2x+1\right)-\sqrt[3]{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt[3]{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-b^3-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x+1}=\sqrt[3]{2x+1}\) (\(x\ge-\frac{1}{2}\))
\(\Leftrightarrow\left(x+1\right)^3=\left(2x+1\right)^2\)
\(\Leftrightarrow x=?\)
2.
ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\Leftrightarrow8x^3+2x-\left(2x+2\right)\sqrt{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-\left(b^2+1\right)b=0\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow2x=\sqrt{2x+1}\) (\(x\ge0\))
\(\Leftrightarrow4x^2=2x+1\)
\(\Leftrightarrow x=?\)
1) đk: \(x\ge1\)
Ta có: \(\sqrt{x-1}-\sqrt{2x\left(x-1\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}=\sqrt{2x\left(x-1\right)}\)
\(\Leftrightarrow x-1=2x^2-2x\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)
Vậy x = 1
2) đk: \(x\ge\frac{1}{2}\)
Ta có: \(\sqrt{5x^2}=2x-1\)
\(\Leftrightarrow5x^2=\left(2x-1\right)^2\)
\(\Leftrightarrow5x^2=4x^2-4x+1\)
\(\Leftrightarrow x^2+4x-1=0\)
\(\Leftrightarrow\left(x+2\right)^2-5=0\)
\(\Leftrightarrow\left(x+2-\sqrt{5}\right)\left(x+2+\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2+\sqrt{5}\left(ktm\right)\\x=-2-\sqrt{5}\left(ktm\right)\end{cases}}\)
=> PT vô nghiệm
3) đk: \(x\ge-1\)
Ta có: \(\sqrt{x+1}+\sqrt{9x+9}=4\)
\(\Leftrightarrow\sqrt{x+1}+3\sqrt{x+1}=4\)
\(\Leftrightarrow4\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=1\)
\(\Rightarrow x=0\)
4) đk: \(x\ge2\)
Ta có: \(\sqrt{x-2}-\sqrt{x\left(x-2\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}=\sqrt{x\left(x-2\right)}\)
\(\Leftrightarrow x-2=x\left(x-2\right)\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
Vậy x = 2
6) đk: \(x\ge-\frac{7}{5}\)
Ta có: \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
\(\Leftrightarrow\frac{2x-3}{x-1}=2\)
\(\Leftrightarrow2x-3=2x-2\)
\(\Leftrightarrow0x=1\) vô lý
=> PT vô nghiệm