1. 

A = \(\frac{x^2-10y^2+10z^2}{y+z}+\frac{y^2-10z^2+10x^2}{z+x}+\frac{z^2-10x^2+10y^2}{x+y}\)

A = \(\frac{x^2}{y+z}-10\frac{\left(y-z\right)\left(y+z\right)}{y+z}+\frac{y^2}{x+z}-10\frac{\left(z-x\right)\left(z+x\right)}{z+x}+\frac{z^2}{x+y}-10\frac{\left(x-y\right)\left(x+y\right)}{x+y}\)

A = \(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}-10y+10z-10z+10x-10x+10y\)

\(A=x\left(\frac{x}{y+z}+1\right)+y\left(\frac{y}{z+x}+1\right)+z\left(\frac{z}{x+y}+1\right)-\left(x+y+z\right)\)

A = \(x\cdot\frac{x+y+z}{y+z}+y\cdot\frac{x+y+z}{z+x}+y\cdot\frac{x+y+z}{x+y}-\left(x+y+z\right)\)

A = \(\left(x+y+z\right)\cdot\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)-\left(x+y+z\right)\)

A = \(\left(x+y+z\right)-\left(x+y+z\right)=0\)

Mn giải giúp e vs ((

Ta có: \(4x^2+2y^2-4xy+4+\sqrt{\left(x+y+z\right)^2}=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+y^2+4+\left|x+y+z\right|=0\)

\(\Leftrightarrow\left(2x-y\right)^2+y^2+\left|x+y+z\right|=-4\)

Mà \(VT\ge0\left(\forall x,y,z\right)\) => vô lý

=> PT vô nghiệm

\(4x^2+2y^2-4xy+4+\sqrt{\left(x+y+z\right)^2}=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+y^2+4+\left|x+y+z\right|=0\)

\(\Leftrightarrow\left(2x-y\right)^2+y^2+\left|x+y+z\right|+4=0\)(1)

Vì \(\left(2x-y\right)^2\ge0\); \(y^2\ge0\); \(\left|x+y+z\right|\ge0\forall x,y,z\)

\(\Rightarrow\left(2x-y\right)^2+y^2+\left|x+y+z\right|\ge0\forall x,y,z\)

\(\Rightarrow\left(2x-y\right)^2+y^2+\left|x+y+z\right|+4\ge4\forall x,y,z\)(2)

Từ (1) và (2) \(\Rightarrow\)Vô lý 

Vậy không tìm được giá trị của x, y, z thỏa mãn đề bài

a) đk: \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

b) Ta có:

\(P=\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{3x-8\sqrt{x}+27}{9-x}\)

\(P=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)+2\sqrt{x}\cdot\left(\sqrt{x}-3\right)-3x+8\sqrt{x}-27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{x+5\sqrt{x}+6+2x-6\sqrt{x}-3x+8\sqrt{x}-27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{7\sqrt{x}-21}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{7\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{7}{\sqrt{x}+3}\)

c) Nếu x không là số chính phương => P vô tỉ (loại)

=> x là số chính phương khi đó để P nguyên thì:

\(\left(\sqrt{x}+3\right)\inƯ\left(7\right)\) , mà \(\sqrt{x}+3\ge3\left(\forall x\ge0\right)\)

\(\Rightarrow\sqrt{x}+3=7\Leftrightarrow\sqrt{x}=4\Rightarrow x=16\)

Vậy x = 16 thì P nguyên

a) \(4\sqrt{\frac{2}{9}}+\sqrt{2}+\sqrt{\frac{1}{18}}\)

\(=\frac{8\sqrt{2}}{6}+\frac{6\sqrt{2}}{6}+\frac{\sqrt{2}}{6}\)

\(=\frac{15\sqrt{2}}{6}=\frac{5\sqrt{2}}{2}\)

b) \(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\)

\(=\frac{\sqrt{3}+1}{3-1}-\frac{\sqrt{3}-1}{3-1}\)

\(=\frac{\sqrt{3}+1-\sqrt{3}+1}{2}=1\)

ai giải giúp mk vs

đag cần gấp

a) Ta có: AB.cosB + cosC.AC=\(\frac{AB^2}{BC}+\frac{AC^2}{BC}\)=\(\frac{BC^2}{BC}\)=BC

b) CMR: tam giác ABC đồng dạng với tam giác AFE(g-g)

\(\Rightarrow\)\(\frac{AB}{AF}=\frac{BC}{EF}\)

\(\Rightarrow\)AB.EF=BC.AF

CMR: tam giác ABH đồng dạng với tam giác AHE (g-g)

\(\Rightarrow\)\(\frac{AB}{AH}=\frac{AH}{AE}\)

\(\Rightarrow\)\(\frac{AH}{AE}=\frac{AH.AB}{AH^2}\)\(\Rightarrow\)\(\frac{AH}{AE}=\frac{EF.AB}{AH^2}\)

\(\Rightarrow\)\(\frac{AH}{AE}=\frac{AF.BC}{AH^2}\)\(\Rightarrow\frac{AH^3}{BC}=AE.AF\)

Ta có:\(S_{AEHF}=AE.AF\)

\(\Rightarrow S_{AEHF}=\frac{AH^3}{BC}\)

\(\frac{5+\sqrt{5}}{\sqrt{5}+2}+\frac{5}{\sqrt{5}-1}-\frac{3\sqrt{5}}{3+\sqrt{5}}\)

\(=\frac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{5-4}+\frac{5\left(\sqrt{5}+1\right)}{5-1}-\frac{3\sqrt{5}\left(3-\sqrt{5}\right)}{9-5}\)

\(=5\sqrt{5}-10+5-2\sqrt{5}+\frac{5\sqrt{5}+5}{4}-\frac{9\sqrt{5}-15}{4}\)

\(=3\sqrt{5}-5+\frac{-4\sqrt{5}+20}{4}\)

\(=3\sqrt{5}-5-\sqrt{5}+5=2\sqrt{5}\)