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Ta có
\(\hept{\begin{cases}2x^2-y^2=1\left(1\right)\\xy+x^2=2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}4x^2-2y^2=2\left(2\right)\\xy+x^2=2\left(3\right)\end{cases}}\)
trừ (2) cho (3) vế với vế ta được
\(3x^2-2y^2-xy=0\Rightarrow3x\left(x-y\right)+2y\left(x-y\right)=0\)
\(\Rightarrow\left(x-y\right)\left(3x+2y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-y=0\\3x+2y=0\end{cases}\Rightarrow\orbr{\begin{cases}x=y\\x=-\frac{2y}{3}\end{cases}}}\)
Sau đó bạn thay vào (1) rồi giải tiếp nhé
1) \(\sqrt{2x-5}=7\)
\(\left(\sqrt{2x-5}\right)^2=7^2\)
\(2x-5=49\)
\(2x=54\)
\(x=27\)
2) \(3+\sqrt{x-2}=4\)
\(\sqrt{x-2}=1\)
\(\left(\sqrt{x-2}\right)^2=1^2\)
\(x-2=1\)
\(x=3\)
1) \(\sqrt{2x-5}=7\left(đk:x\ge\dfrac{5}{2}\right)\)
\(\Leftrightarrow2x-5=49\Leftrightarrow2x=54\Leftrightarrow x=27\left(tm\right)\)
2) \(3+\sqrt{x-2}=4\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)
3) \(\Leftrightarrow\sqrt{\left(x-1\right)^2}=1\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
4) \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\Leftrightarrow\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
5) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x+4\right)^2}\)
\(\Leftrightarrow\left|2x-1\right|=\left|x+4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
6) \(ĐK:x\ge-2\)
\(\Leftrightarrow5\sqrt{x+2}-3\sqrt{x+2}-\sqrt{x+2}=\sqrt{x+7}\)
\(\Leftrightarrow\sqrt{x+2}=\sqrt{x+7}\)
\(\Leftrightarrow x+2=x+7\Leftrightarrow2=7\left(VLý\right)\)
Vậy \(S=\varnothing\)
7) \(ĐK:x\ge-1\)
\(\Leftrightarrow5\sqrt{2x+1}+3\sqrt{x+1}=4\sqrt{x+1}+4\sqrt{2x+1}\)
\(\Leftrightarrow\sqrt{2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow2x+1=x+1\Leftrightarrow x=0\left(tm\right)\)
\(a,\sqrt{2x-1}=2\)
\(\Rightarrow2x-1=4\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=\frac{5}{2}\)
\(b,\sqrt{2x-1}=x+1\)
\(\Rightarrow2x-1=\left(x+1\right)^2\)
\(\Rightarrow2x-1=x^2+2x+1\)
\(\Rightarrow x^2+2x-2x=-1-1\)
\(\Rightarrow x^2=-2VN\)
Giải giúp e vs ạ ,e đang cần gấp
Chứng minh 2 ý đó giúp e vs ạ e đang cần gấp
ta có :
\(\frac{1}{cos^2x}=\frac{sin^2x+cos^2x}{cos^2x}=1+\left(\frac{sinx}{cosx}\right)^2=1+tan^2x\)
\(\frac{1}{sin^2x}=\frac{sin^2x+cos^2x}{sin^2x}=1+\left(\frac{cosx}{sinx}\right)^2=1+cot^2x\)
Bài 1:
1: \(\sqrt{3+2\sqrt{2}}=\sqrt{2}+1\)
2: \(\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}\)
3: \(\sqrt{11-2\sqrt{30}}=\sqrt{6}-\sqrt{5}\)
4: \(\sqrt{7-2\sqrt{10}}=\sqrt{5}-\sqrt{2}\)
\(x+y=2\Rightarrow y=2-x\)
\(xy=x.\left(2-x\right)=2x-x^2=-\left(x^2-2x\right)\)
\(=-\left(x^2-2x+1-1\right)=-\left(x-1\right)^2+1=1-\left(x-1\right)^2\le1\)
=> đpcm
( Dấu "=" xảy ra <=> x = 1 => y = 2 - x = 2 - 1 = 1 )
Bạn có cách giải bằng hình ko ạ