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1: Ta có: \(\sqrt{3x-5}=2\)
\(\Leftrightarrow3x-5=4\)
hay x=3
2: Ta có: \(\sqrt{25\left(x-1\right)}=20\)
\(\Leftrightarrow x-1=16\)
hay x=17
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài I:
a: \(=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{x+y+2\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{x-y}{\sqrt{x}+\sqrt{y}}\left(x,y>0\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)\)
\(=2\sqrt{y}\)
\(B=\dfrac{x+y-2\sqrt{xy}}{\sqrt{x}-\sqrt{y}}-\dfrac{x-y}{\sqrt{x}+\sqrt{y}}\left(x,y>0\right)\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\left(\sqrt{x}-\sqrt{y}\right)-\left(\sqrt{x}-\sqrt{y}\right)\)
\(=0\)
\(C=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}+\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}\sqrt{y}}\left(x,y>0\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}+\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}=\sqrt{x}+\sqrt{y}+\sqrt{x}+\sqrt{y}\)
\(=2\left(\sqrt{x}+\sqrt{y}\right)\)
\(D=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\dfrac{y\sqrt{x}-x\sqrt{y}}{\sqrt{xy}}\left(x,y>0\right)\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{xy}\left(\sqrt{y}-\sqrt{x}\right)}{\sqrt{xy}}=\sqrt{x}-\sqrt{y}+\sqrt{y}-\sqrt{x}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi thời gian làm riêng của người 1 và người 2 lần lượt là a,b
Theo đề, ta có:
8/a+8/b=1/2 và 3/a+6/b=1/4
=>a=24 và b=48
![](https://rs.olm.vn/images/avt/0.png?1311)
f: \(F=\dfrac{1-\sqrt{x-1}}{\sqrt{x-2\sqrt{x-1}}}\)
\(=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1}-1}\)
=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
`1a)A=4/(sqrt5-2)+\sqrt{(sqrt5-2)^2}-sqrt{125}`
`=(4(sqrt5+2))/(5-4)+sqrt5-2-5sqrt5`
`=4(sqrt5+2)-4sqrt5-2`
`=4sqrt5+8-4sqrt5-2`
`=6`
`B=(1/(sqrtx-2)+sqrtx/(x-4)).(3x-12)/(2sqrtx+2)`
`đk:x>=0,x ne 4`
`B((sqrtx+2+sqrtx)/(x-4)).(3(x-4))/(2sqrtx+2)`
`=(2sqrtx+2)/(x-4).(3(x-4))/(2sqrtx+2)`
`=3`
Câu 7:
Đặt A=\(\sqrt{a^2+abc}+\sqrt{b^2+abc}+\sqrt{c^2+abc}\)
\(=\sqrt{a}\sqrt{a+bc}+\sqrt{b}\sqrt{b+ac}+\sqrt{c}\sqrt{c+ab}\)\(\le\sqrt{\left(a+b+c\right)\left(a+b+c+ab+bc+ac\right)}\) (theo bđt bunhia)
\(\Rightarrow A\le\sqrt{1+ab+bc+ac}\)
mà \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\) (bạn tự chứng minh được)
\(\Rightarrow A\le\sqrt{1+\dfrac{\left(a+b+c\right)^2}{3}}=\sqrt{1+\dfrac{1}{3}}=\dfrac{2\sqrt{3}}{3}\)
Áp dụng bđt cosi có:
\(1=a+b+c\ge\sqrt[3]{abc}\) \(\Leftrightarrow abc\le\dfrac{1}{27}\)
Có \(M=A+9\sqrt{abc}\le\dfrac{2\sqrt{3}}{3}+9\sqrt{\dfrac{1}{27}}=\dfrac{5\sqrt{3}}{3}\)
=> maxM\(=\dfrac{5\sqrt{3}}{3}\) \(\Leftrightarrow a=b=c=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: Để hai đường thẳng song song thì \(\left\{{}\begin{matrix}m^2=4\\m\ne2\end{matrix}\right.\Leftrightarrow m=-2\)
b: Để hai đường thẳng vuông góc thì \(4m^2=-1\)(vô lý)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: Để hàm số nghịch biến thì \(2m-1< 0\)
hay \(m< \dfrac{1}{2}\)