a. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)

<=> \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

<=> \(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

<=> \(P=\dfrac{\sqrt{x}+2}{x-2\sqrt{x}}\)

b. Khi \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) => \(\sqrt{x}=2+\sqrt{3}\)

=> \(P=\dfrac{2+\sqrt{3}+2}{7+4\sqrt{3}-2\left(2+\sqrt{3}\right)}=\dfrac{4+\sqrt{3}}{7+4\sqrt{3}-4-2\sqrt{3}}=\dfrac{4+\sqrt{3}}{3+2\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)

check giùm mik

Bài 1: 

a: \(\sqrt{0.49a^2}=-0.7a\)

b: \(\sqrt{25\left(a-7\right)^2}=5a-35\)

c: \(\sqrt{a^4\left(a-2\right)^2}=a^2\cdot\left(a-2\right)\)

d: \(\dfrac{1}{a-3b}\cdot\sqrt{a^6\left(a-3b\right)^2}\)

\(=\dfrac{1}{a-3b}\cdot a^3\cdot\left(a-3b\right)=a^3\)

Bài 2: 

a: \(2\left(x+y\right)\cdot\sqrt{\dfrac{1}{x^2+2xy+y^2}}\)

\(=2\left(x+y\right)\cdot\dfrac{1}{x+y}\)

=2

b: \(\dfrac{3x}{7y}\cdot\sqrt{\dfrac{49y^2}{9x^2}}\)

\(=\dfrac{3x}{7y}\cdot\dfrac{-7y}{3x}\)

=-1

\(1,\\ a,=\dfrac{\left(3+2\sqrt{3}\right)\sqrt{3}}{3}+\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{1}\\ =\dfrac{3\sqrt{3}+6}{3}+\sqrt{2}=\sqrt{3}+1+\sqrt{2}\\ b,=\left(\dfrac{\sqrt{5}+\sqrt{2}}{3}-\dfrac{\sqrt{5}-\sqrt{2}}{3}+1\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}+3}{3}\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{2\sqrt{2}+3}{3\left(3+2\sqrt{2}\right)}=\dfrac{1}{3}\)

\(2,\\ A=2x+\sqrt{\left(x-3\right)^2}=2x+\left|x-3\right|\\ =2\left(-5\right)+\left|-5-3\right|=-10+8=-2\\ B=\dfrac{\sqrt{\left(2x+1\right)^2}}{\left(x-4\right)\left(x+4\right)}\left(x-4\right)^2=\dfrac{\left|2x+1\right|\left(x-4\right)}{x+4}\\ B=\dfrac{17\cdot4}{12}=\dfrac{17}{3}\)

Tối thiểu em phải ghi đúng đề ra chứ. Đường cao là đường cao nào? H là điểm nào? Đó là những chi tiết trong đề còn thiếu

dạ e cảm ơn ạ
hồi nãy e vội quá nên quên ghi ạ

hình bé quá

sin 65⁰=cos 35⁰
\(cos70^0=sin30^0\)
\(tan80^0=cot20^0\)
\(cot68^0=tan32^0\)

Bài 3:

\(a,=\sqrt[3]{\left(x-1\right)^3}-\sqrt[3]{\left(5x+1\right)^3}=x-1-5x-1=-4x-2\\ b,=6a-6a+20a=20a\)

Bài 2:

\(a,=2\sqrt[3]{6}+3\sqrt[3]{5}-4\sqrt[3]{6}-2\sqrt[3]{5}=\sqrt[3]{5}-2\sqrt[3]{6}\\ b,=\sqrt[3]{8}-4\sqrt[3]{27}+2\sqrt[3]{64}=2-12+16=6\\ c,=\sqrt[3]{64}+\sqrt[3]{48}+\sqrt[3]{36}-\sqrt[3]{48}-\sqrt[3]{36}-\sqrt[3]{27}=4-3=1\\ d,=\sqrt[3]{162\left(-2\right)\cdot\dfrac{2}{3}}=\sqrt[3]{-216}=-6\)

Thank you

a) để \(\sqrt{4-2x}\) có nghĩa thì

\(4-2x\text{≥}0\)

⇒\(4\text{≥}2x\)

⇒\(2\text{≥}x\)

b) để \(\sqrt{\dfrac{-3}{2x+3}}\) có nghĩa thì

\(\dfrac{-3}{2x+3}\text{≥}0\)

⇒\(2x+3< 0\)

⇒\(2x< -3\)

⇒\(x< -\dfrac{3}{2}\)

Bài 3: 

a: Ta có: \(C=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

\(=a+\sqrt{a}-2\sqrt{a}-1+1\)

\(=a-\sqrt{a}\)

b: Để C=2 thì \(\sqrt{a}-2=0\)

hay a=4