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\(\sqrt{x-4\sqrt{x-1}+3}+\sqrt{x-6\sqrt{x-1}+8}=1\\ < =>\sqrt{x-1-2\sqrt{x-1}.2+4}+\sqrt{x-1-2\sqrt{x-1}.3+9}=1\\ < =>\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)ĐK: x>=1
\(< =>|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1\\ < =>\left(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\right)^2=1\\ < =>\sqrt{x-1}-2+2\left|\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}-3\right)\right|+\sqrt{x-1}-3=1\\ < =>2\sqrt{x-1}-5+2\left|x+5-5\sqrt{x-1}\right|=1\\ < =>2\left|x+5-5\sqrt{x-1}\right|=6-2\sqrt{x-1}\\ < =>\left|x+5-5\sqrt{x-1}\right|=3-\sqrt{x-1}\)
\(< =>\left[{}\begin{matrix}x+5-5\sqrt{x-1}=3-\sqrt{x-1}\left(1\right)\\x+5-5\sqrt{x-1}=\sqrt{x-1}-3\left(2\right)\end{matrix}\right.\)
Giải (1): \(x+5-5\sqrt{x-1}=3-\sqrt{x-1}\\ < =>x+2-4\sqrt{x-1}=0\\ < =>x-1-2\sqrt{x-1}.2+4=1\\ < =>\left(\sqrt{x-1}-2\right)^2=1\\ < =>\left[{}\begin{matrix}\sqrt{x-1}-2=1\\\sqrt{x-1}-2=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=8\\x=0\left(loại\right)\end{matrix}\right.\)
Giải (2) cũng ra x=8
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1.\(A=\left(\sqrt{3}+1\right)\sqrt{\dfrac{14-6\sqrt{3}}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\dfrac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\dfrac{44\left(2-\sqrt{3}\right)}{22}}=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
2.1.a) \(x^2=\left(x-1\right)\left(3x-2\right)\Leftrightarrow x^2=3x^2-5x+2\Leftrightarrow2x^2-5x+2=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
b) \(9x^4+5x^2-4=0\Leftrightarrow9x^4+9x^2-4x^2-4=0\)
\(\Leftrightarrow9x^2\left(x^2+1\right)-4\left(x^2+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(9x^2-4\right)=0\)
mà \(x^2+1>0\Rightarrow9x^2=4\Rightarrow x^2=\dfrac{4}{9}\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2) Gọi số xe lúc đầu của đội là a(xe) \(\left(a\in N,a>0\right)\)
Theo đề,ta có: \(\left(a-2\right)\left(\dfrac{120}{a}+3\right)=120\Leftrightarrow120+3a-\dfrac{240}{a}-6=120\)
\(\Leftrightarrow\dfrac{3a^2-6a-240}{a}=0\Rightarrow3a^2-6a-240=0\Rightarrow a^2-2a-80=0\)
\(\Leftrightarrow\left(a+8\right)\left(a-10\right)=0\) mà \(a>0\Rightarrow a=10\)
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Lời giải:
$\Delta'=(\sqrt{3})^2-(\sqrt{3}-1)(1-\sqrt{3})=7-2\sqrt{3}$
PT có 2 nghiệm:
\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{\sqrt{3}+\sqrt{7-2\sqrt{3}}}{1-\sqrt{3}}\)
\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{\sqrt{3}-\sqrt{7-2\sqrt{3}}}{1-\sqrt{3}}\)
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2:
\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)
\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)
B=(x1+x2)^2-2x1x2
=3^2-2*(-7)
=9+14=23
C=căn (x1+x2)^2-4x1x2
=căn 3^2-4*(-7)=căn 9+28=căn 27
D=(x1^2+x2^2)^2-2(x1x2)^2
=23^2-2*(-7)^2
=23^2-2*49=431
D=9x1x2+3(x1^2+x2^2)+x1x2
=10x1x2+3*23
=69+10*(-7)=-1
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\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\Leftrightarrow\sqrt{\left(2-\sqrt{x-1}\right)^2}+\sqrt{\left(3+\sqrt{x-1}\right)^2}=5\)
\(\Leftrightarrow|2-\sqrt{x-1}|+3+\sqrt{x-1}=5\)
\(\Leftrightarrow\orbr{\begin{cases}2-\sqrt{x-1}+\sqrt{x-1}=2\\\sqrt{x-1}-2+\sqrt{x-1}=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}1\le x\le5\\x=5\end{cases}}\)
\(\Rightarrow1\le x\le5\)
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cách khác đơn giản hơn nhiều
Đk:\(x\ge1\)
\(pt\Leftrightarrow\sqrt{2\left(x-1\right)\left(x+4\right)}+\sqrt{2\left(x-1\right)\left(x+3\right)}-3\sqrt{x+4}-3\sqrt{x+3}-1=0\)
\(\Leftrightarrow\sqrt{2\left(x-1\right)\left(x+4\right)}-3\sqrt{x+4}+\sqrt{2\left(x-1\right)\left(x+3\right)}-3\sqrt{x+3}=1\)
\(\Leftrightarrow\sqrt{x+4}\left(\sqrt{2\left(x-1\right)}-3\right)+\sqrt{x+3}\left(\sqrt{2\left(x-1\right)}-3\right)=1\)
\(\Leftrightarrow\left(\sqrt{x+4}+\sqrt{x+3}\right)\left(\sqrt{2\left(x-1\right)}-3\right)=1\)
Xét Ư(1)={1;-1}={....}
Dễ nhé, tự làm nốt
Đk: \(x\ge1\)
\(pt\Leftrightarrow\sqrt{2x^2+6x-8}+\sqrt{2x^2+4x-6}-3\sqrt{x+4}-3\sqrt{x+3}-1=0\)
\(\Leftrightarrow\sqrt{2x^2+6x-8}-\frac{10}{3}\sqrt{x+3}+\frac{1}{3}\sqrt{x+3}-1\sqrt{2x^2+4x-6}-3\sqrt{x+4}=0\)
\(\Leftrightarrow\frac{2x^2+6x-8-\frac{100}{9}\left(x+3\right)}{\sqrt{2x^2+6x-8}+\frac{10}{3}\sqrt{x+3}}+\frac{x-6}{3\left(\sqrt{x+3}+3\right)}+\frac{2x^2+4x-6-9\left(x+4\right)}{\sqrt{2x^2+4x-6}+3\sqrt{x+4}}=0\)
Để đỡ rối ta đặt mấy cái mẫu \(\hept{\begin{cases}N=\sqrt{2x^2+6x-8}+\frac{10}{3}\sqrt{x+3}>0\\H=\sqrt{x+3}+3>0\\T=\sqrt{2x^2+4x-6}+3\sqrt{x+4}>0\end{cases}}\)
\(\Leftrightarrow\frac{18x^2-46x-372}{9N}+\frac{x-6}{3H}+\frac{2x^2-5x-42}{T}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{18x+62}{9N}+\frac{1}{3H}+\frac{2x+7}{T}\right)=0\)
Dễ thấy: \(\forall x\ge1\) thì \(\frac{18x+62}{9N}+\frac{1}{3H}+\frac{2x+7}{T}>0\)
\(\Rightarrow x-6=0\Rightarrow x=6\) (thỏa mãn)
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