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Ta có: \(\left(2x-3\right)^2-\left(x-3\right)\left(x^2-5x+4\right)\)
\(=4x^2-12x+9-x^3+5x^2-4x+3x^2-15x+12\)
\(=-x^3+12x^2-31x+21\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Do x>=17 nên |x-17|>=0
Do đó |x-17|=x-17
=>A=x-17+18-x
=>A=1
Chúc em học tốt^^
Khi \(x\ge17\)thì \(x-17\ge0\)hay \(\left|x-17\right|=x-17\)
\(\Rightarrow A=x-17+18-x=1\)
tk nka !!11
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2\\ P=\left(x-y-x-y\right)^2-4x^2\\ P=4y^2-4x^2=4\left(y-x\right)\left(x+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{40x-7-17}{5x-3}=\dfrac{40x-24}{5x-3}=\dfrac{8\left(5x-3\right)}{5x-3}=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\sqrt{27}-\sqrt{\dfrac{16}{3}}-\sqrt{48}-\sqrt{8\dfrac{1}{3}}\)
\(=6\sqrt{3}-4\sqrt{\dfrac{1}{3}}-4\sqrt{3}-5\sqrt{\dfrac{1}{3}}\)
\(=2\sqrt{3}-9\sqrt{\dfrac{1}{3}}\)
\(=2\sqrt{3}-3\sqrt{9\cdot\dfrac{1}{3}}\)
\(=2\sqrt{3}-3\sqrt{3}\)
\(=-\sqrt{3}\)
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\(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\)
\(=\left(5\sqrt{5}-2\sqrt{3}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+3\sqrt{3}\right)\)
\(=\left(3\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}+2\sqrt{3}\right)\)
\(=\left(3\sqrt{5}\right)^2-\left(2\sqrt{3}\right)^2\)
\(=15-12\)
\(=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17=12x-12\)
5(x+2)(x−2)−(2x−3)2−x2+17(x2−4)−(4x2−12x+9)−x2+17=12x−12
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}\)
\(\Leftrightarrow A^2=18-2\sqrt{\left(9+\sqrt{17}\right)\left(9-\sqrt{17}\right)}\)
\(=18-2\sqrt{81-17}=2\)
\(\Rightarrow A=\sqrt{2}\)
\(\Rightarrow C=A-\sqrt{2}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Không nhé bạn, vì 48/17 là số tối giản rồi !
nhầm, là phân số tối giản