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pt <=> \(\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(1+\sqrt{3}\right)\)
\(-3\left(x-1\right)\left(x-\sqrt{3}\right)\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}\right)\)
\(+2\left(x-1\right)\left(x-\sqrt{2}\right)\left(\sqrt{3}+1\right)\left(\sqrt{3}+\sqrt{2}\right)=3x-1\)
<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left[\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)-\left(x-1\right)\left(\sqrt{2}+\sqrt{3}\right)\right]\)
\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left[\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)-\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)\right]\)
\(=3x-1\)
<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(x+\sqrt{3}\right)\left(1-\sqrt{2}\right)\)
\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left(x+1\right)\left(\sqrt{2}-\sqrt{3}\right)=3x-1\)
<=> \(3-x^2-2\left(1-x^2\right)=3x-1\)
<=> \(x^2-3x+2=0\) phương trình bậc 2.
Em làm tiếp nhé!
\(a,Đk:1\le x\le4\)
Đặt \(y=\sqrt{4-x}+\sqrt{2x-2}\)Ta có: \(y^2=4-x+2x-2+2\sqrt{\left(4-x\right)\left(2x-2\right)}\)
\(\Leftrightarrow x+2+2\sqrt{\left(4-x\right)\left(2x-2\right)}=y^2\Leftrightarrow x+2\sqrt{\left(4-x\right)\left(2x-2\right)}=y^2-2\)
Phương trình trở thành: \(5+y^2-2=4y\)
\(\Leftrightarrow y^2-4y+3=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=3\end{cases}}\) ( Vì \(a+b+c=0\))
- \(y=1.\) Ta có: \(\sqrt{4-x}+\sqrt{2x-2}=1\Leftrightarrow\sqrt{2x-2}=1-\sqrt{4-x}\)
\(\Leftrightarrow\hept{\begin{cases}1-\sqrt{4-x}\ge0\\2x-2=\left(1-\sqrt{4-x}\right)^2\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le1\\2x-2=1-2\sqrt{4-x}+4-x\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}0\le4-x\le1\\2\sqrt{4-x}=7-3x\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}3\le x\le4;7-3x\ge0\\4\left(4-x\right)=\left(7-3x\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in\varnothing\\4\left(4-x\right)=\left(7-3x\right)^2\end{cases}}\) \(\Leftrightarrow x\in\varnothing\)
- \(y=3\)Ta có: \(\sqrt{4-x}+\sqrt{2x-2}=3\Leftrightarrow\sqrt{2x-2}=3-\sqrt{4-x}\)
\(\Leftrightarrow\hept{\begin{cases}3-\sqrt{4-x}\ge0\\2x-2=\left(3-\sqrt{4-x}\right)^2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le3\\2x-2=9-6\sqrt{4-x}+4-x\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le3\\2\sqrt{4-x}=5-x\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}0\le4-x\le9;5-x\ge0\\4\left(4-x\right)=\left(5-x\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-5\le x\le4\\x^2-6x+9=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}-5\le x\le4\\\left(x-3\right)^2=0\end{cases}}\Leftrightarrow x=3\)
Vậy pt có nghiệm duy nhất là \(x=3\)
(Làm xong hoa mắt :((
dk \(\hept{\begin{cases}x\left(3x+1\right)\ge0\\x\left(x-1\right)\ge0\end{cases}< =>\orbr{\begin{cases}x\ge1\\x\le\frac{-1}{3}\end{cases}}}\)
vì x khác 0 nên chia cả 2 vế cho \(\sqrt{x}\)ta được \(\sqrt{3x+1}-\sqrt{x-1}=2\sqrt{x}< =>\)\(\sqrt{x-1}+2\sqrt{x}-\sqrt{3x+1}=0< =>\)\(\sqrt{x-1}+\frac{4x-\left(3x+1\right)}{2\sqrt{x}+\sqrt{3x+1}}=0\)\(\sqrt{x-1}+\frac{x-1}{2\sqrt{x}+\sqrt{3x+1}}=0\)\(< =>\sqrt{x-1}\left(1+\frac{\sqrt{x-1}}{2\sqrt{x}+\sqrt{3x+1}}\right)=0< =>\sqrt{x-1}=0\) (vì biểu thức trong ngoặc luôn \(\ge1\)) <=> x-1= 0 <=> x=1 (thỏa mãn điều kiện)
1:
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-2\right)=0\)
=>x-3=0 hoặc \(\sqrt{x+3}=2\)
=>x=3 hoặc x+3=4
=>x=1(loại) hoặc x=3(nhận)
2:
\(\Leftrightarrow\left(\sqrt{4x+1}-\sqrt{3x-4}\right)^2=1\)
=>\(4x-1+3x-4-2\sqrt{\left(4x+1\right)\left(3x-4\right)}=1\)
=>\(\sqrt{4\left(4x+1\right)\left(3x-4\right)}=7x-6\)
=>4(12x^2-16x+3x-4)=(7x-6)^2
=>49x^2-84x+36=48x^2-52x-16
=>-84x+36=-52x-16
=>-32x=-52
=>x=13/8
3: =>\(\sqrt{\left(x-5\right)^2}=5-x\)
=>|x-5|=5-x
=>x-5<=0
=>x<=5
4: \(\Leftrightarrow\left|x-4\right|=x+2\)
=>\(\left\{{}\begin{matrix}x>=-2\\\left(x-4\right)^2=\left(x+2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\x^2-8x+16=x^2+4x+4\end{matrix}\right.\)
=>x>=-2 và -8x+16=4x+4
=>x=1
\(a,\left(x^2-4x+11\right)\left(x^4-8x^2+21\right)=35\)
Phương trình trên tương đương với:
\(\left[\left(x-2\right)^2+7\right]\left[\left(x^2-4\right)^2+5\right]=35\left(1\right)\)
Do: \(\hept{\begin{cases}\left(x-2\right)^2+7\ge7\forall x\\\left(x^2-4\right)^2+5\ge5\forall x\end{cases}}\Rightarrow\left[\left(x+2\right)^2+7\right]\left[\left(x^2+4\right)^2+5\right]\ge35\forall x\)
Nên: \(\left(1\right)\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2+7=7\\\left(x^2-4\right)^2+5=5\end{cases}\Leftrightarrow}x=2\)
Vậy ..................................
\(b,\sqrt{x}+\sqrt{1-x}+\sqrt{x\left(1-x\right)}=1\)
\(Đkxđ:0\le x\le1\) Đặt: \(0< a=\sqrt{x}+\sqrt{1-x}\Rightarrow\frac{a^2-1}{2}=\sqrt{x\left(1-x\right)}\)
\(+)\) Phương trình mới là: \(a+\frac{a^2-1}{2}=1\Leftrightarrow a^2+2a-3=0\Leftrightarrow\left(a-1\right)\left(a+3\right)=0\)
\(\Leftrightarrow a=\left\{-3;1\right\}\Rightarrow a=1>0\)
\(\sqrt{x}+\sqrt{1-x}=1\)
\(+)\) Nếu \(a=1\Leftrightarrow x+1-x+2\sqrt{x\left(1-x\right)}=1\Leftrightarrow\sqrt{x\left(1-x\right)}=0\)
\(\Rightarrow x=\left\{0;1\right\}\left(tm\right)\)
Vậy .............................