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Câu 1: ĐK: x khác -1/2, y khác -2
Đặt \(\sqrt[3]{\frac{2x+1}{y+2}}=t\) Từ phương trình thứ nhất ta có:
\(t+\frac{1}{t}=2\Leftrightarrow t^2-2t+1=0\Leftrightarrow t=1\)
=> \(\sqrt[3]{\frac{2x+1}{y+2}}=1\Leftrightarrow2x+1=y+2\Leftrightarrow2x-y=1\)
Vậy nên ta có hệ phương trình cơ bản: \(\hept{\begin{cases}2x-y=1\\4x+3y=7\end{cases}}\)Em làm tiếp nhé>
\(1,ĐKXĐ:\hept{\begin{cases}y\ne-2\\x\ne-\frac{1}{2}\end{cases}}\)
Đặt \(\sqrt[3]{\frac{2x+1}{y+2}}=a\left(a\ne0\right)\)
\(Pt\left(1\right)\Leftrightarrow a+\frac{1}{a}=2\)
\(\Leftrightarrow a^2+1=2a\)
\(\Leftrightarrow\left(a-1\right)^2=0\)
\(\Leftrightarrow a=1\)
\(\Leftrightarrow\sqrt[3]{\frac{2x+1}{y+2}}=1\)
a)\(3x^2+6x-3=\sqrt{\frac{x+7}{3}}\)
Đk:\(x\ge-7\)
\(pt\Leftrightarrow9x^4+36x^3+18x^2-36x+9=\frac{x+7}{3}\)
\(\Leftrightarrow9x^4+36x^3+18x^2-36x+9-\frac{x+7}{3}=0\)
\(\Leftrightarrow\left(x^2+\frac{5x}{3}-\frac{4}{3}\right)\left(9x^2+21x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{\sqrt{69}+7}{6}\\x=\frac{\sqrt{73}-5}{6}\end{cases}}\) (thỏa)
b)\(2x^2+2x+1=\left(2x+3\right)\left(\sqrt{x^2+x+2}-1\right)\)
\(\Leftrightarrow2x^2+2x+1=\left(2x+3\right)\sqrt{x^2+x+2}-2x-3\)
\(\Leftrightarrow2x^2+4x+4=\left(2x+3\right)\sqrt{x^2+x+2}\)
\(\Leftrightarrow\frac{2x^2+4x+4}{2x+3}=\sqrt{x^2+x+2}\)
\(\Leftrightarrow\frac{2x^2+4x+4}{2x+3}-2x=\sqrt{x^2+x+2}-2x\)
\(\Leftrightarrow\frac{2x^2+4x+4}{2x+3}-2x=\frac{x^2+x+2-4x^2}{\sqrt{x^2+x+2}+2x}\)
\(\Leftrightarrow\frac{-2\left(x+2\right)\left(x-1\right)\left(3x+2\right)}{\left(2x+3\right)\left(3x+2\right)}=\frac{x^2+x+2-4x^2}{\sqrt{x^2+x+2}+2x}\)
\(\Leftrightarrow\frac{-2\left(x+2\right)\left(x-1\right)\left(3x+2\right)}{\left(2x+3\right)\left(3x+2\right)}=\frac{-\left(x-1\right)\left(3x+2\right)}{\sqrt{x^2+x+2}+2x}\)
\(\Leftrightarrow\frac{-2\left(x+2\right)\left(x-1\right)\left(3x+2\right)}{\left(2x+3\right)\left(3x+2\right)}-\frac{-\left(x-1\right)\left(3x+2\right)}{\sqrt{x^2+x+2}+2x}=0\)
\(\Leftrightarrow-\left(x-1\right)\left(3x+2\right)\left(\frac{2\left(x+2\right)}{\left(2x+3\right)\left(3x+2\right)}-\frac{1}{\sqrt{x^2+x+2}+2x}\right)=0\)
\(\Leftrightarrow x=1;x=-\frac{2}{3}\) (thỏa)
Câu 1:
PT \(\Leftrightarrow x^2+3x+8=(x+5)\sqrt{x^2+x+2}\)
\(\Leftrightarrow (x^2+x+2)+2(x+5)-4=(x+5)\sqrt{x^2+x+2}\)
Đặt \(\sqrt{x^2+x+2}=a; x+5=b(a\geq 0)\)
\(PT\Leftrightarrow a^2+2b-4=ba\)
\(\Leftrightarrow (a^2-4)-b(a-2)=0\)
\(\Leftrightarrow (a-2)(a+2-b)=0\Rightarrow \left[\begin{matrix} a=2\\ a+2=b\end{matrix}\right.\)
Nếu \(a=2\Rightarrow x^2+x+2=a^2=4\)
\(\Leftrightarrow x^2+x-2=0\Leftrightarrow (x-1)(x+2)=0\Rightarrow x=1; x=-2\) (đều thỏa mãn)
Nếu \(a+2=b\Leftrightarrow \sqrt{x^2+x+2}+2=x+5\)
\(\Leftrightarrow \sqrt{x^2+x+2}=x+3\)
\(\Rightarrow \left\{\begin{matrix} x+3\geq 0\\ x^2+x+2=(x+3)^2\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+3\geq 0\\ 5x+7=0\end{matrix}\right.\Rightarrow x=\frac{-7}{5}\) (thỏa mãn)
Vậy..........
Câu 2:
ĐKXĐ: \(x\geq 1\) hoặc \(x\leq \frac{1}{2}\)
\(10x^2-9x-8x\sqrt{2x^2-3x+1}+3=0\)
\(\Leftrightarrow 3(2x^2-3x+1)-8x\sqrt{2x^2-3x+1}+4x^2=0\)
Đặt \(\sqrt{2x^2-3x+1}=a(a\geq 0)\)
Khi đó PT \(\Leftrightarrow 3a^2-8xa+4x^2=0\)
\(\Leftrightarrow (a-2x)(3a-2x)=0\) \(\Rightarrow \left[\begin{matrix} a=2x\\ 3a=2x\end{matrix}\right.\)
Nếu \(a=\sqrt{2x^2-3x+1}=2x\Rightarrow \left\{\begin{matrix} x\geq 0\\ 2x^2-3x+1=4x^2\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x\geq 0\\ 2x^2+3x-1=0\end{matrix}\right.\Rightarrow x=\frac{-3+\sqrt{17}}{4}\) (t/m)
Nếu \(3a=3\sqrt{2x^2-3x+1}=2x\Rightarrow \left\{\begin{matrix} x\geq 0\\ 9(2x^2-3x+1)=4x^2\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x\geq 0\\ 14x^2-27x+9=0\end{matrix}\right.\Rightarrow x=\frac{3}{2}; x=\frac{3}{7}\) (t/m)
Vậy...........
a)Đk:\(x\ge\frac{1}{2}\)
\(pt\Leftrightarrow4x^2-12x+4+4\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(2x-1\right)^2-4\left(2x-1\right)-1+4\sqrt{2x-1}=0\)
Đặt \(t=\sqrt{2x-1}>0\Rightarrow\hept{\begin{cases}t^2=2x-1\\t^4=\left(2x-1\right)^2\end{cases}}\)
\(t^4-4t^2+4t-1=0\)
\(\Leftrightarrow\left(t-1\right)^2\left(t^2+2t-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}t-1=0\\t^2+2t-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}t=1\\t=\sqrt{2}-1\end{cases}\left(t>0\right)}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=2-\sqrt{2}\end{cases}}\) là nghiệm thỏa pt
a/ ĐKXĐ:...
\(\Leftrightarrow4x^2-4x\sqrt{2x-1}-3x^2+6x-3=0\)
\(\Leftrightarrow4x\left(x-\sqrt{2x-1}\right)-3\left(x-1\right)^2=0\)
\(\Leftrightarrow\frac{4x\left(x-1\right)^2}{x+\sqrt{2x-1}}-3\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\frac{4x}{x+\sqrt{2x-1}}=3\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4x=3x+3\sqrt{2x-1}\)
\(\Leftrightarrow x=3\sqrt{2x-1}\)
\(\Leftrightarrow x^2-18x+9=0\) \(\Rightarrow9\pm6\sqrt{2}\)
Vậy pt có 3 nghiệm....
b/ ĐKXĐ:...
\(\Leftrightarrow4x^2-4x\sqrt{4x-3}-x^2+4x-3=0\)
\(\Leftrightarrow4x\left(x-\sqrt{4x-3}\right)-\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\frac{4x\left(x^2-4x+3\right)}{x+\sqrt{4x-3}}-\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\Rightarrow x=...\\\frac{4x}{x+\sqrt{4x-3}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4x=x+\sqrt{4x-3}\)
\(\Leftrightarrow3x=\sqrt{4x-3}\)
\(\Leftrightarrow9x^2-4x+3=0\) (vô nghiệm)
Vậy...
Lời giải:
Đặt \(\sqrt{x+2}=a(a\geq 0)\Rightarrow 2=a^2-x\)
Khi đó pt đã cho trở thành:
\(x^3-3x^2+2a^3-3x.2=0\)
\(\Leftrightarrow x^3-3x^2+2a^3-3x(a^2-x)=0\)
\(\Leftrightarrow x^3+2a^3-3xa^2=0\)
\(\Leftrightarrow x(x^2-a^2)-2a^2(x-a)=0\)
\(\Leftrightarrow (x-a)(x^2+xa-2a^2)=0\)
\(\Leftrightarrow (x-a)[(x^2-a^2)+a(x-a)]=0\)
\(\Leftrightarrow (x-a)^2(x+2a)=0\)
TH1: \(x-a=0\Rightarrow x=a=\sqrt{x+2}\Rightarrow \left\{\begin{matrix} a\geq 0\\ x^2=x+2\end{matrix}\right.\)
\(\Rightarrow x=2\)
TH2: \(x+2a=0\Rightarrow x=-2a=-2\sqrt{x+2}\)
\(\Rightarrow \left\{\begin{matrix} x\leq 0\\ x^2=4(x+2)\end{matrix}\right.\Rightarrow x=2-2\sqrt{3}\)
Vậy PT có nghiệm \(x\in \left\{2-2\sqrt{3}; 2\right\}\)
sao thầy(cô) trả lời nhanh quá vậy sao em trả lời kịp
SP không cánh mà đi