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\(4\sqrt{2}x^2-6x-\sqrt{2}=0\) \(0\)
\(\left(a=4\sqrt{2};b=-6;b'=-3;c=-\sqrt{2}\right)\)
\(\Delta'=b'^2-ac\)
\(=\left(-3\right)^2-4.\left(-\sqrt{2}\right)\)
\(=9+4\sqrt{2}\)
\(\sqrt{\Delta}=\sqrt{9+4\sqrt{2}}\)
Vay : phương trình có 2 nghiệp phân biệt
\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{3+\sqrt{9+4\sqrt{2}}}{4\sqrt{2}}\)
\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{3-\sqrt{9+4\sqrt{2}}}{4\sqrt{2}}\)
Bài 1:
Căn bậc hai số học của \(\left(-7\right)^2\) là |-7|=7
Bài 2:
a: \(0,2\cdot\sqrt{\left.\left(-10\right)^2\right.\cdot3}+2\cdot\sqrt{\left(\sqrt5-\sqrt3\right)^2}\)
\(=0,2\cdot10\cdot\sqrt3+2\cdot\left(\sqrt5-\sqrt3\right)\)
\(=2\sqrt3+2\sqrt5-2\sqrt3=2\sqrt5\)
Bài 3:
\(\sqrt{\left(2x-1\right)^2}-5=0\)
=>\(\left|2x-1\right|-5=0\)
=>|2x-1|=5
=>\(\left[\begin{array}{l}2x-1=5\\ 2x-1=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=6\\ 2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-2\end{array}\right.\)
Câu 4:
ĐKXĐ: 4-3x>=0
=>3x<=4
=>\(x\le\frac43\)
Câu 6:
\(\sin^6\alpha+cos^6\alpha+3\cdot\sin^2\alpha\cdot cos^2\alpha\)
\(=\left(\sin^2\alpha+cos^2\alpha\right)^3-3\cdot\sin^2\alpha\cdot cos^2\alpha\cdot\left(\sin^2\alpha+cos^2\alpha\right)+3\cdot\sin^2\alpha\cdot cos^2\alpha\)
\(=1-3\cdot\sin^2\alpha\cdot cos^2\alpha+3\cdot\sin^2\alpha\cdot cos^2\alpha\)
=1
Câu 2:
\(\left(2\sqrt3-3\sqrt2\right)^2+2\sqrt6+3\sqrt{24}\)
\(=12+18-2\cdot2\sqrt3\cdot3\sqrt2+2\sqrt6+3\cdot2\sqrt6\)
\(=30-12\sqrt6+2\sqrt6+6\sqrt6=30-4\sqrt6\)
(Với x > 0; x 1; x4)
\(a,\)\(\sqrt{x^2-2x+1}=\sqrt{\left(x-1\right)^2}\)
\(đkxđ\Leftrightarrow\sqrt{\left(x-1\right)^2}\ge0\)
\(\Rightarrow x-1\ge0\Rightarrow x\ge1\)
\(b,\)\(\sqrt{x+3}+\sqrt{x+9}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x+3\ge0\\x+9\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge-3\\x\ge-9\end{cases}}}\)
\(\Rightarrow x\ge-3\)
\(c,\)\(\sqrt{\frac{x-1}{x+2}}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x+2\ne0\\\frac{x-1}{x+2}\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-2\\\frac{x-1}{x+2}\ge0\end{cases}}}\)
\(\frac{x-1}{x+2}\ge0\)\(\Rightarrow\orbr{\begin{cases}x-1\ge0;x+2>0\\x-1\le0;x+2< 0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x\ge-1;x>-2\\x\le1;x< 2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x\ge-1\\x< 2\end{cases}}\)
Vậy căn thức xác định khi x \(\ge\)-1 hoawck x < 2
1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)
\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)
\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)
\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)
\(=-8\sqrt{3}\)
2) \(A=\sqrt{12-4x}\) có nghĩa khi:
\(12-4x\ge0\)
\(\Leftrightarrow4x\le12\)
\(\Leftrightarrow x\le\dfrac{12}{4}\)
\(\Leftrightarrow x\le3\)
3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)
biểu thức e viết liền quá khó phân biệt ví dụ như x +1 -\(\frac{2\sqrt{x}}{\sqrt{x-1}}\)hay là x +\(\frac{1-\sqrt{2x}}{\sqrt{x-1}}\)