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Câu 1:
\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}=2\left(x+1\right)\)
- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) pt vô nghiệm
- Nhận thấy \(x=-1\) là 1 nghiệm
- Nếu \(x>-1\) kết hợp ĐKXĐ các căn thức ta được \(x\ge1\), pt tương đương:
\(\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\)
\(\Leftrightarrow2x+6+x-1+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4x+4\)
\(\Leftrightarrow2\sqrt{2x^2+4x-6}=x-1\)
\(\Leftrightarrow4\left(2x^2+4x-6\right)=\left(x-1\right)^2\)
\(\Leftrightarrow7x^2+18x-25=0\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{25}{7}< 0\left(l\right)\end{matrix}\right.\)
Vậy pt có nghiệm \(x=\pm1\)
Câu 2:
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1+2\sqrt{x-1}+1}-\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|=2\)
- Nếu \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\) pt trở thành:
\(\sqrt{x-1}+1-\sqrt{x-1}+1=2\Leftrightarrow2=2\) (luôn đúng)
- Nếu \(1\le x< 2\) pt trở thành:
\(\sqrt{x-1}+1-1+\sqrt{x-1}=2\Leftrightarrow x=2\left(l\right)\)
Vậy nghiệm của pt là \(x\ge2\)
Câu 3:
Bình phương 2 vế ta được:
\(2x^2+2x+5+2\sqrt{\left(x^2+x+4\right)\left(x^2+x+1\right)}=2x^2+2x+9\)
\(\Leftrightarrow\sqrt{\left(x^2+x+4\right)\left(x^2+x+1\right)}=2\)
\(\Leftrightarrow\left(x^2+x+4\right)\left(x^2+x+1\right)=4\)
Đặt \(x^2+x+1=a>0\) pt trở thành:
\(a\left(a+3\right)=4\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Câu 5:
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
Mà \(VT=\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1\)
\(\Rightarrow VT\ge VP\Rightarrow\) Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\le0\end{matrix}\right.\) \(\Rightarrow5\le x\le10\)
Vậy nghiệm của pt là \(5\le x\le10\)
a) \(\sqrt{x-1}+\sqrt{2x-1}=5\)
\(\Leftrightarrow3x-2+2\sqrt{\left(x-1\right)\left(2x-1\right)}=25\)
\(\Leftrightarrow2\sqrt{\left(x-1\right)\left(2x-1\right)}=25-3x+2\)
\(\Leftrightarrow2\sqrt{\left(x-1\right)\left(2x-1\right)}=-3x+27\)
Bình phương 2 vế, ta được:
\(\Leftrightarrow4\left(x-1\right)\left(2x-1\right)=9\left(x-9\right)^2\)
\(\Leftrightarrow8x^2-4x-8x+4=9x^2-162x+729\)
\(\Leftrightarrow8x^2-12x+4-9x^2+162x-729=0\)
\(\Leftrightarrow-x^2+150x-725=0\)
\(\Leftrightarrow x^2-150x+725=0\)
\(\Leftrightarrow\left(x-145\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-145=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=145\left(ktm\right)\\x=5\left(tm\right)\end{cases}}\)
\(\Rightarrow x=5\)
b) \(x+\sqrt{2x-1}-2=0\)
\(\Leftrightarrow\sqrt{2x-1}=2-x\)
Bình phương 2 vế, ta được:
\(\Leftrightarrow2x-1=4-4x^2+x^2=0\)
\(\Leftrightarrow2x-1-4+4x-x^2=0\)
\(\Leftrightarrow6x-5-x^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)
a)\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=3\)
\(\Leftrightarrow\left|1-x\right|+\left|x-2\right|=3\)
Có: \(VT=\left|1-x\right|+\left|x-2\right|\)
\(\ge\left|1-x+x-2\right|=3=VP\)
Khi \(x=0;x=3\)
b)\(\sqrt{x^2-10x+25}=3-19x\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=3-19x\)
\(\Leftrightarrow\left|x-5\right|=3-19x\)
\(\Leftrightarrow x^2-10x+25=361x^2-114x+9\)
\(\Leftrightarrow-360x^2+104x+16=0\)
\(\Leftrightarrow-5\left(5x-2\right)\left(9x+1\right)=0\)
\(\Rightarrow x=\frac{2}{5};x=-\frac{1}{9}\)
c)\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)
\(\Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=5\)
\(\Leftrightarrow2\sqrt{2x-3}+5=5\)\(\Leftrightarrow\sqrt{2x-3}=0\Leftrightarrow x=\frac{3}{2}\)
Bài 1:
a/ ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}-2+\sqrt{2x-1}-3=0\)
\(\Leftrightarrow\frac{x-5}{\sqrt{x-1}+2}+\frac{2\left(x-5\right)}{\sqrt{2x-1}+3}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{\sqrt{x-1}+2}+\frac{2}{\sqrt{2x-1}+3}\right)=0\)
\(\Rightarrow x=5\)
b/ĐKXĐ:...
\(x-1+\sqrt{2x-1}-1=0\)
\(\Leftrightarrow x-1+\frac{2\left(x-1\right)}{\sqrt{2x-1}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(1+\frac{2}{\sqrt{2x-1}+1}\right)=0\)
\(\Rightarrow x=1\)
Bài 2:
\(A=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)
\(B=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left(3-\sqrt{6}\right)+\left(2\sqrt{6}-3\right)\)
\(=\sqrt{6}\)
\(C=\left(\frac{3+\sqrt{5}-3+\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right).\frac{\left(\sqrt{5}-1\right)}{\sqrt{5}\left(\sqrt{5}-1\right)}\)
\(=\frac{2\sqrt{5}}{4}.\frac{1}{\sqrt{5}}=\frac{1}{2}\)
\(\sqrt{x^2-2x+1}\) + \(\sqrt{x^2-4x+4}\) = 3
<=> \(\sqrt{\left(x-1\right)^2}\)+ \(\sqrt{\left(x-2\right)^2}\)= 3
<=> \(\left|x-1\right|\)+\(\left|x-2\right|\)=3
<=> x - 1 + x - 2 = 3
<=> 2x - 3 = 3
<=> x = \(\dfrac{6}{2}\)= 3
b ,
\(\sqrt{x^2-10x+25}=3-19x\)
<=>\(\sqrt{\left(x-5\right)^2}=3-19x\)
<=> \(\left|x-5\right|=3-19x\)
<=> \(x-5=3-19x\)
\(\Leftrightarrow x+19x=3+5\)
\(\Leftrightarrow20x=8\Leftrightarrow x=\dfrac{8}{20}=\dfrac{2}{5}\)
b: \(\Leftrightarrow\left(x^2+5x+4\right)=5\sqrt{x^2+5x+28}\)
Đặt \(x^2+5x+4=a\)
Theo đề, ta có \(5\sqrt{a+24}=a\)
=>25a+600=a2
=>a=40 hoặc a=-15
=>x2+5x-36=0
=>(x+9)(x-4)=0
=>x=4 hoặc x=-9
c: \(\Leftrightarrow x^2+5x=2\sqrt[3]{x^2+5x-2}-2\)
Đặt \(x^2+5x=a\)
Theo đề, ta có: \(a=2\sqrt[3]{a}-2\)
\(\Leftrightarrow\sqrt[3]{8a}=a+2\)
=>(a+2)3=8a
=>\(a^3+6a^2+12a+8-8a=0\)
\(\Leftrightarrow a^3+6a^2+4a+8=0\)
Đến đây thì bạn chỉ cần bấm máy là xong
Bài 1:
a/ \(\sqrt{\dfrac{2x^2-4x+2}{6}}=1\) .
\(\Leftrightarrow\dfrac{2\left(x^2-2x+1\right)}{6}=1\)
\(\Leftrightarrow\dfrac{\left(x-1\right)^2}{3}=1\)
\(\Leftrightarrow\left(x-1\right)^2=3\) \(\Rightarrow\left[{}\begin{matrix}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}+1\\x=-\sqrt{3}+1\end{matrix}\right.\)
vậy tập nghiệm của phương trình S=\(\left\{1-\sqrt{3};\sqrt{3}+1\right\}\)
b/ ta có: \(\dfrac{6}{x-4}=\sqrt{2}\Leftrightarrow\sqrt{2}\left(x-4\right)=6\)
\(\Leftrightarrow x\sqrt{2}-4\sqrt{2}=6\)
\(\Leftrightarrow x\sqrt{2}=6+4\sqrt{2}\)
\(\Leftrightarrow x=\dfrac{6+4\sqrt{2}}{2}=4+3\sqrt{2}\)
vậy \(x=4+3\sqrt{2}\) là nghiệm của phương trình
c/ \(\sqrt{\dfrac{20}{2x^2-8x+8}}=\sqrt{5}\)
\(\Leftrightarrow\left(\sqrt{\dfrac{20}{2x^2-8x+8}}\right)^2=\left(\sqrt{5}\right)^2\)
\(\Leftrightarrow\dfrac{20}{2\left(x^2-4x+4\right)}=5\)
\(\Leftrightarrow\dfrac{10}{\left(x-2\right)^2}=\dfrac{10}{2}\)
\(\Rightarrow\left(x-2\right)^2=2\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{2}\\x-2=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{2}\\x=2-\sqrt{2}\end{matrix}\right.\)
vậy tập nghiệm của phương trình \(S=\left\{2+\sqrt{2};2-\sqrt{2}\right\}\)
Bài 2:
a/ đặt A= \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow A^2=3+\sqrt{5}+3-\sqrt{5}-2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(\Leftrightarrow A^2=6-2\sqrt{9-5}\)
\(\Leftrightarrow A^2=6-2\sqrt{4}=6-4=2\)
\(\Rightarrow A=\sqrt{2}\)
\(\Rightarrow\)\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\) = \(\sqrt{2}\)
\(\Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
b/ \(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\dfrac{\sqrt{12}}{\sqrt{15}}+\dfrac{\sqrt{75}}{\sqrt{15}}+\dfrac{\sqrt{27}}{\sqrt{15}}=\sqrt{\dfrac{12}{15}}+\sqrt{\dfrac{75}{15}}+\sqrt{\dfrac{27}{15}}\)
\(=\dfrac{2\sqrt{5}}{5}+\sqrt{5}+\dfrac{3\sqrt{5}}{5}=\left(\dfrac{2\sqrt{5}}{5}+\dfrac{3\sqrt{5}}{5}\right)+\sqrt{5}\)
\(=\sqrt{5}+\sqrt{5}=2\sqrt{5}\)
c/ \(\left(12\sqrt{20}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(24\sqrt{5}-80\sqrt{2}+105\sqrt{2}\right):\sqrt{10}\)
\(=\left(24\sqrt{5}+25\sqrt{2}\right):\sqrt{10}=\dfrac{24\sqrt{5}}{\sqrt{10}}+\dfrac{25\sqrt{2}}{\sqrt{10}}\)
\(=12\sqrt{2}+5\sqrt{5}\)
Đặt \(2x-5=t^2\)ta có \(x=\frac{t^2+5}{2}\)thay giá trị của x vào phương trình đã cho được:
\(\sqrt{\frac{t^2+5}{2}-2+t}+\sqrt{\frac{t^2+5}{2}+2+3t}=7\sqrt{2}\)
hay \(\sqrt{t^2+5-2+2t}+\sqrt{t^2+5+4+6t}=14\)
\(\sqrt{t^2+2t+1}+\sqrt{t^2+6t+9}=14\)
\(\sqrt{\left(t+1\right)^2}+\sqrt{\left(t+3\right)^2}=14\)
\(t+1+t+3=14\)
\(2t+4=14\)
2t=10
t=5
Từ đó \(x=\frac{25+5}{2}=15\)
\(\sqrt{13}-\sqrt{12}=\frac{1}{\sqrt{13}+\sqrt{12}}\) ; \(\sqrt{7}-\sqrt{6}=\frac{1}{\sqrt{7}+\sqrt{6}}\)
Mà \(\sqrt{13}+\sqrt{12}>\sqrt{7}+\sqrt{6}\Rightarrow\frac{1}{\sqrt{13}+\sqrt{12}}< \frac{1}{\sqrt{7}+\sqrt{6}}\)
\(\Rightarrow\sqrt{13}-\sqrt{12}< \sqrt{7}-\sqrt{6}\)
ĐKXĐ: \(x\ge\frac{5}{2}\)
\(\Leftrightarrow\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)
\(\Leftrightarrow\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)
\(\Leftrightarrow\left|\sqrt{2x-5}+3\right|+\left|\sqrt{2x-5}-1\right|=4\)
\(\Leftrightarrow\left|\sqrt{2x-5}+3\right|+\left|1-\sqrt{2x-5}\right|=4\)
Mà \(\left|\sqrt{2x-5}+3\right|+\left|1-\sqrt{2x-5}\right|\ge\left|\sqrt{2x-5}+3+1-\sqrt{2x-5}\right|=4\)
Dấu "=" xảy ra khi và chỉ khi \(1-\sqrt{2x-5}\ge0\Rightarrow2x-5\le1\Rightarrow x\le3\)
Vậy nghiệm của pt là \(\frac{5}{2}\le x\le3\)