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Bài 1:
Đk:\(x\ge\frac{1}{2}\)
Đặt \(\sqrt{2x-1}=t\Rightarrow2x=t^2+1\)
\(pt\Leftrightarrow\left(t^2+1\right)^2-8\left(t^2+4\right)t=7-22\left(t^2+1\right)\)
\(\Leftrightarrow t^4-8t^3+24t^2-32t+16=0\)
\(\Leftrightarrow\left(t-2\right)^4=0\Leftrightarrow t=2\Leftrightarrow\sqrt{2x-1}=2\)
\(\Leftrightarrow2x-1=4\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\) (thỏa mãn)
Bài 2:
Cộng 2 vế với \(7x^2+23x+12\) ta được:
\(\left(x+2\right)^3+\left(x+2\right)=\left(7x^2+23x+12\right)+\sqrt[3]{7x^2+23x+12}\)
\(\Leftrightarrow\left(x+2\right)^3=7x^2+23x+12\)
\(\Leftrightarrow x^3+6x^2+12x+8=7x^2+23x+12\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+3x+1\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=4\\x=\frac{\sqrt{5}-3}{2}\end{matrix}\right.\) (thỏa mãn)
\(1\))\(x^2+5x+8=3\sqrt{x^3+5x^2+7x+6}\left(1\right)\\ĐK:x\ge-\dfrac{3}{2} \\ \left(1\right)\Leftrightarrow x^2+5x+8=3\sqrt{\left(2x+3\right)\left(x^2+x+2\right)}\left(2\right)\)
Đặt \(b=\sqrt{2x+3};a=\sqrt{x^2+x+2}\)
\(\left(2\right)\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)\(\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1\pm\sqrt{5}}{2}\\x=\dfrac{7\pm\sqrt{89}}{2}\end{matrix}\right.\)
4)\(ĐK:x\ge-\dfrac{1}{3}\)
\(x^2-7x+2+2\sqrt{3x+1}=0\\ \Leftrightarrow x^2-7x+6+2\sqrt{3x+1}-4=0\\ \Leftrightarrow\left(x-1\right)\left(x-6\right)+\dfrac{12\left(x-1\right)}{2\sqrt{3x+1}+4}=0\\ \Leftrightarrow\left(x-1\right)\left(x-6+\dfrac{12}{2\sqrt{3x+1}+4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x-6+\dfrac{12}{2\sqrt{3x+1}+4}=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(x-5\right)+\dfrac{6}{\sqrt{3x+1}+2}-1=0\\ \Leftrightarrow\left(x-5\right)+\dfrac{4-\sqrt{3x+1}}{\sqrt{3x+1}+2}=0\\ \Leftrightarrow\left(x-5\right)-\dfrac{3\left(x-5\right)}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}=0\\ \Leftrightarrow\left(x-5\right)\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)=3\\ \Leftrightarrow3x+1+6\sqrt{3x+1}+8=3\\ \Leftrightarrow x+2\sqrt{3x+1}+2=0\\ \Leftrightarrow2\sqrt{3x+1}=-x-2\ge0\Leftrightarrow x\le-2\)
Vậy pt có 2 nghiệm là x=1 và x=5
Mình giải mẫu pt đầu thôi nhé, những pt sau ttự.
1,\(x^4-\frac{1}{2}x^3-x^2-\frac{1}{2}x+1=0\)
Ta thấy x=0 ko là nghiệm.
Chia cả 2 vế cho x2 >0:
pt\(\Leftrightarrow x^2-\frac{1}{2}x-1-\frac{1}{2x}+\frac{1}{x^2}=0\)
Đặt \(t=x-\frac{1}{x}\left(t\in R\right)\)
\(\Rightarrow x^2+\frac{1}{x^2}=t^2+2\)
pt\(\Leftrightarrow t^2-\frac{1}{2}t+1=0\)(vô n0)
Vậy pt vô n0.
#Walker
1) Phương trình đã cho tương đương
\(\Leftrightarrow\left(x-2\right)\left(3\sqrt{x^2+1}-x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=\frac{3}{4}\end{matrix}\right.\)
2.
\(DK:\hept{\begin{cases}x\ge-\frac{1}{5}\\x\ne0\end{cases}}\)
PT
\(\Leftrightarrow6+3\sqrt{5x+1}\left(\sqrt{5x+1}-1\right)=14\left(\sqrt{5x+1}-1\right)\)
\(\Leftrightarrow15x+23-17\sqrt{5x+1}=0\)
\(\Leftrightarrow\left(68-17\sqrt{5x+1}\right)+\left(15x-45\right)=0\)
\(\Leftrightarrow\frac{17\left(x-3\right)}{4+\sqrt{5x+1}}+15\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{17}{4+\sqrt{5x+1}}+15\right)=0\)
Vi \(\frac{17}{4+\sqrt{5x+1}}+15>0\)
\(\Rightarrow x=3\left(n\right)\)
Vay nghiem cua PT la \(x=3\)
đặt \(\sqrt{x^2+x+1}=t\left(t\ge\sqrt{\dfrac{3}{4}}\right)tacó\)
pt \(\Leftrightarrow\)3t=t\(^2\)+2
\(\Leftrightarrow\)\(\left[{}\begin{matrix}t=1\left(tm\right)\\t=2\left(tm\right)\end{matrix}\right.\)
Với t=1 ta có x\(^2\)+x+1=1 \(\Leftrightarrow\)x=0 hoặc x=-1
với t=2 ta có x\(^2\)+x+1 =2 \(\Leftrightarrow\)\(\dfrac{-1\mp\sqrt{5}}{2}\)=x
câu 2 tương tự đặt 2x^2+x-2=t(t\(\ge\dfrac{-17}{8}\))
ta có pt \(\Leftrightarrow\)t^2+5t-6=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}t=1\left(tm\right)\\t=-6\left(loại\right)\end{matrix}\right.\)
với t=1 thì 2x^2+x-2=1 \(\Leftrightarrow\)t=1 hoặc -3/2