Rút gọn biểu thức hộ mk nhé
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)\left(x+2\right)+\left(x+1\right)x=x^2+2x-x-2+x^2+x=\left(x^2+x^2\right)+\left(2x-x+x\right)-2=2x^2+2x-2=2\left(x^2+x-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét 2 trường hợp :
TH1 : x ≥ 0 => |x - 3| = x - 3
=> 3(x - 1) - 2|x - 3| = 3(x - 1) - 2(x - 3)
= 3x - 3 - 2x + 6
= x + 3
TH2 : x < 0 => |x - 3| = 3 - x
=> 3(x - 1) - 2|x - 3| = 3(x - 1) - 2(3 - x)
= 3x - 3 - 6 + 2x
= 5x - 9
Vậy 3(x - 1) - 2|x - 3| = x + 3 hoặc 3(x - 1) - 2|x - 3| = 5x - 9
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{3\cdot5\cdot7\cdot11\cdot13\cdot37-10101}{1212120+40404}\)
\(=\frac{(3\cdot7\cdot11\cdot13\cdot37)\cdot5-10101\cdot1}{120\cdot10101+4\cdot10101}\)
\(=\frac{10101\cdot(5-1)}{10101\cdot(120+4)}=\frac{4}{124}=\frac{1}{31}\)
~Chúc bạn học tốt~
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)
\(=\)\(2\sqrt{3}.\sqrt{3}+\sqrt{5}.\sqrt{3}-\sqrt{60}\)
\(=\)\(6+\sqrt{15}-\sqrt{60}\)
\(=\)\(6-\sqrt{15}\)
\(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)
=\(6+\sqrt{15}-\sqrt{2^2.15}\)
=\(6+\sqrt{15}-2\sqrt{15}\)
=\(6-\sqrt{15}\)
k mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
ĐKXĐ: $x>0; x\neq 1$
\(P=\frac{1}{\sqrt{x}+1}+\frac{x}{\sqrt{x}(1-\sqrt{x})}=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\)
\(=\frac{1-\sqrt{x}+\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(1-\sqrt{x})}=\frac{x+1}{1-x}\)
b. Khi $x=\frac{1}{\sqrt{2}}$ thì:
\(P=\frac{\frac{1}{\sqrt{2}}+1}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{2}\)