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Akai Haruma, No choice teen, Arakawa Whiter, HISINOMA KINIMADO, tth, Nguyễn Việt Lâm, Phạm Hoàng Lê Nguyên, @Nguyễn Thị Ngọc Thơ
Mn giúp em vs ạ! Thanks trước!
Ta có: \(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}+1}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
f) ĐKXĐ: \(x\ge-\frac{3}{2}\)
Khi đó VT > 0 nên \(VT>0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\left(L\right)\end{matrix}\right.\)
Lũy thừa 6 cả 2 vế lên PT tương đương:
\( \left( x-3 \right) \left( {x}^{11}+9\,{x}^{10}+6\,{x}^{9}-142\,{x}^{ 8}-231\,{x}^{7}+1113\,{x}^{6}+2080\,{x}^{5}-4604\,{x}^{4}-6908\,{x}^{3 }+13222\,{x}^{2}+10983\,x-15327 \right) =0\)
Cái ngoặc to vô nghiệm vì nó tương đương:
\(\left( x-2 \right) ^{11}+31\, \left( x-2 \right) ^{10}+406\, \left( x -2 \right) ^{9}+2906\, \left( x-2 \right) ^{8}+12281\, \left( x-2 \right) ^{7}+31031\, \left( x-2 \right) ^{6}+46656\, \left( x-2 \right) ^{5}+46648\, \left( x-2 \right) ^{4}+46452\, \left( x-2 \right) ^{3}+44590\, \left( x-2 \right) ^{2}+36015\,x-55223 = 0\)(vô nghiệm với mọi \(x\ge2\))
Vậy x = 3.
PS: Nghiệm đẹp thế này chắc có cách AM-Gm độc đáo nhưng mình chưa nghĩ ra
@Akai Haruma, @Nguyễn Việt Lâm
giúp em vs ạ! Cần gấp ạ
em cảm ơn nhiều!
\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\\ \)(1)
\(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\\ \)
\(x^2+1\ge1\forall x\Rightarrow2x+1\ge0\Rightarrow!2x+1!=2x+1\)
\(\left(1\right)\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\\ \)
\(\left(1\right)\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\\ \)
\(\left(1\right)\Leftrightarrow2x+1=\left(2x+1\right)\left(x^2+1\right)\Leftrightarrow\left(2x+1\right).\left(1-\left(x^2+1\right)\right)=0\)
\(\left\{\begin{matrix}2x+1=0\\-x^2=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-\frac{1}{2}\\x=0\end{matrix}\right.\)
\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)
\(\Leftrightarrow\sqrt{\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[2\left(x+\frac{1}{2}\right)\left(x^2+1\right)\right]\)
\(\Leftrightarrow\sqrt{\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)}=\left(x+\frac{1}{2}\right)\left(x^2+1\right)\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)\left(x-\frac{1}{2}+1\right)}-\left(x+\frac{1}{2}\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)}-\left(x+\frac{1}{2}\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}-\left(x+\frac{1}{2}\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)-\left(x+\frac{1}{2}\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)\left(-1-x^2+1\right)=0\)
\(\Leftrightarrow-x^2\left(x+\frac{1}{2}\right)=0\)\(\Leftrightarrow\left[\begin{matrix}-x^2=0\\x+\frac{1}{2}=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=0\\x=-\frac{1}{2}\end{matrix}\right.\)
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
con 6 tách trong căn thành nhân tử nhân 2 vế cho 2 rồi tách thành hđt
c) Ta có:
\(\sqrt{x+\frac{3}{x}}=\frac{x^2+7}{2\left(x+1\right)}\)
\(\Leftrightarrow\sqrt{x+\frac{3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)
\(\Leftrightarrow\frac{\sqrt{x^2+3}-2\sqrt{x}}{\sqrt{x}}=\frac{x^2-4x+3}{2\left(x+1\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{\sqrt{x^3+3x}+2x}=\frac{x^2-4x+3}{2\left(x+1\right)}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4x+3=0\\\sqrt{x^3+3x}+2x=2\left(x+1\right)\end{cases}}\)
+) \(x^2-4x+3=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
+) \(\sqrt{x^3+3x}+2x=2x+2\Rightarrow x=1\)
a/ Đặt \(\sqrt{2\left(x^2-x\right)}=a\)
\(\Rightarrow a^4-2a^2=a\)
\(\Leftrightarrow a\left(a+1\right)\left(a^2-a-1\right)=0\)