Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 3 \(\hept{\begin{cases}x+y+xy=2+3\sqrt{2}\\x^2+y^2=6\end{cases}}\)
\(\hept{\begin{cases}\left(x+y\right)+xy=2+3\sqrt{2}\\\left(x+y\right)^2-2xy=6\end{cases}}\)
\(\hept{\begin{cases}S+P=2+3\sqrt{2}\left(1\right)\\S^2-2P=6\left(2\right)\end{cases}}\)
Từ (1)\(\Rightarrow P=2+3\sqrt{2}-S\)Thế P vào (2) rồi giải tiếp nhé. Mình lười lắm ^.^
a.
\(DK:49-28x-4x^2\ge0\)
PT\(\Leftrightarrow\sqrt{49-28x-4x^2}=5\)
\(\Leftrightarrow49-28x-4x^2=25\)
\(\Leftrightarrow4x^2+28x-24=0\)
\(\Leftrightarrow x^2+7x-6=0\)
Ta co:
\(\Delta=7^2-4.1.\left(-6\right)=73>0\)
\(\Rightarrow\hept{\begin{cases}x_1=\frac{-7+\sqrt{73}}{2}\left(n\right)\\x_2=\frac{-7-\sqrt{73}}{2}\left(n\right)\end{cases}}\)
Vay nghiem cua PT la \(\hept{\begin{cases}x_1=\frac{-7+\sqrt{73}}{2}\\x_2=\frac{-7-\sqrt{73}}{2}\end{cases}}\)
\(\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}=0\)
\(\Leftrightarrow\left(\sqrt[3]{x+1}+1\right)+\sqrt[3]{x+2}+\left(\sqrt[3]{x+3}-1\right)=0\)
\(\Leftrightarrow\frac{x+2}{\sqrt[3]{\left(x+1\right)^2}-\sqrt[3]{x+1}+1}+\frac{x+2}{\sqrt[3]{\left(x+2\right)^4}}+\frac{x+2}{\sqrt[3]{\left(x+3\right)^2}+\sqrt[3]{x+3}+1}\)(liên hợp tử mẫu)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{\sqrt[3]{\left(x+1\right)^2}-\sqrt[3]{x+1}+1}+\frac{1}{\sqrt[3]{\left(x+2\right)^4}}+\frac{1}{\sqrt[3]{\left(x+3\right)^2}+\sqrt[3]{x+3}+1}\right)=0\)
\(\Leftrightarrow x+2=0\)( vì biểu thức thứ 2 luôn khác 0)
\(\Leftrightarrow x=-2\)
Vậy...
\(\left(\sqrt[3]{x+1}+\sqrt[3]{x+3}\right)\left(LH\right)=\sqrt[3]{x+2}\left(LH\right)\)
\(\Leftrightarrow2\left(x+2\right)=\sqrt[3]{x+2}\left(Lh\right)\)
=> x=-2 la nghiệm
x khác -2
\(2\sqrt[3]{\left(x+2\right)^2}=-\left(LH\right)\) Vô nghiệm
a,\(\sqrt{1-x}=\sqrt[3]{27}\left(đk:x\le1\right)\Leftrightarrow\sqrt{1-x}=3\)
\(< =>\sqrt{1-x}^2=9< =>1-x=9< =>x=-8\)tm
b,\(\sqrt{x^2-10x+25}=x+1\)
\(< =>\sqrt{\left(x-5\right)^2}=x+1\)
\(< =>|x-5|=x+1\)
\(< =>\orbr{\begin{cases}-x+5=x+1\left(x< 5\right)\\x-5=x+1\left(x\ge5\right)\end{cases}}\)
\(< =>\orbr{\begin{cases}2x=4< =>x=2\left(tm\right)\\-5-1=0\left(vo-li\right)\end{cases}}\)
c, Đặt \(\sqrt{x}=t\left(t\ge0\right)\)khi đó pt tương đương
\(t^2+t-6=0< =>t^2-2t+3t-6=0\)
<\(< =>t\left(t-2\right)+3\left(t-2\right)=0< =>\left(t+3\right)\left(t-2\right)=0\)
\(< =>\orbr{\begin{cases}t+3=0\\t-2=0\end{cases}}< =>\orbr{\begin{cases}t=-3\left(ktm\right)\\t=2\left(tm\right)\end{cases}}\)
khi đó ta được \(\sqrt{x}=t< =>x=4\)
a) \(\sqrt{1-x}=\sqrt[3]{27}\)
\(\Leftrightarrow\sqrt{1-x}=3\)
\(\Leftrightarrow1-x=9\)
\(\Rightarrow x=-8\)
b) \(\sqrt{x^2-10x+25}=x+1\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x+1\)
\(\Leftrightarrow\left|x-5\right|=x+1\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=x+1\\x-5=-x-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}0=6\left(vl\right)\\2x=4\end{cases}}\Rightarrow x=2\)
c) \(x+\sqrt{x}-6=0\)
\(\Leftrightarrow\left(x+3\sqrt{x}\right)-\left(2\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+3\right)-2\left(\sqrt{x}+3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=-3\left(vl\right)\end{cases}}\Rightarrow x=4\)
\(x^2-15x-6\sqrt{x-1}+74=0\)
\(\Leftrightarrow\left(\left(x-1\right)-6\sqrt{x-1}+9\right)+\left(x^2-16x+64\right)+2=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-3\right)^2+\left(x-8\right)^2+2=0\)
Ta có VT > 0; VP = 0 nên pt vô nghiệm