ĐKXĐ: tự làm.

Rút gọn:

\(A=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(x+\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{0+x-\sqrt{x}}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

\(\left(-x\right)\left(x^2-x+1\right)+\dfrac{1}{2}x^2\left(2x-4\right)+x\cdot x+x\cdot1-2\)

\(=-x^3+x^2-x+x^3-2x^2+x^2+x-2\)

\(=\left(-x+x^3\right)+\left(x^2-2x^2+x^2\right)+\left(-x+x\right)+\left(-2\right)\)

\(=-2\)

Sai rồi nhé. phải là 22kg

+) \(VT=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(=abc+ac^2+b^2c+bc^2+a^2b+a^2c+ab^2+abc\)

\(=a^2b+a^2c+ab^2+2abc+ac^2+b^2c+bc^2\)

+) \(VP=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(=a^2b+abc+a^2c+ab^2+b^2c+abc+abc+bc^2+ac^2+\left(-abc\right)\)

\(=a^2b+a^2c+ab^2+2abc+ac^2+b^2c+bc^2\)

\(\Rightarrow VT=VP\) (đpcm)

Nếu giảm đường kính hình tròn đi 20% thì bán kính hình tròn đó cũng giảm đi 20%.

\(100\%\cdot100\%-80\%\cdot80\%=36\%\)

Diện tích hình tròn là :

\(113,04\div36\cdot100=314\left(cm^2\right)\)

ĐS: \(314cm^2\)

\(\left(2x-5\right)^2-64\)

\(=4x^2-20x-39\)

\(=\left(2x+3\right)\left(2x-13\right)\)

a) \(x^2-25=x^2-5^2=\left(x+5\right)\left(x-5\right)\)

b) \(x^2-x^4=x^2\left(-x+1\right)\left(x+1\right)\)

Gọi số đo góc B là x, số đo góc C là y, số đo góc A là z.

\(7x=13y\Rightarrow\dfrac{x}{13}=\dfrac{y}{7}\)

Ta có: \(x+y+z=180^o\) (định lí tổng 3 góc trong một tam giác)

Mà \(z=90^o\) \(\Rightarrow x+y=90^o\)

- Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{13}=\dfrac{y}{7}\Rightarrow\dfrac{x+y}{13+7}=\dfrac{90}{20}=\dfrac{9}{2}\)

+) \(\dfrac{x}{13}=\dfrac{9}{2}\Rightarrow x=58,5\)

+) \(\dfrac{y}{7}=\dfrac{9}{2}\Rightarrow y=31,5\)

Vậy \(\widehat{B}=58,5^o;\widehat{C}=31,5^o\)

a)

\(M=\dfrac{1}{7+\dfrac{1}{5+\dfrac{1}{3+\dfrac{1}{2}}}}+\dfrac{1}{9+\dfrac{1}{8+\dfrac{1}{7+\dfrac{1}{6}}}}\)

\(=\dfrac{1}{7+\dfrac{1}{5+\dfrac{1}{\dfrac{7}{2}}}}+\dfrac{1}{9+\dfrac{1}{8+\dfrac{1}{\dfrac{43}{6}}}}\)

\(=\dfrac{1}{7+\dfrac{1}{5+\dfrac{2}{7}}}+\dfrac{1}{9+\dfrac{1}{8+\dfrac{6}{43}}}\)

\(=\dfrac{1}{7+\dfrac{1}{\dfrac{37}{7}}}+\dfrac{1}{9+\dfrac{1}{\dfrac{350}{43}}}\)

\(=\dfrac{1}{7+\dfrac{7}{37}}+\dfrac{1}{9+\dfrac{43}{350}}\)

\(=\dfrac{1}{\dfrac{266}{37}}+\dfrac{1}{\dfrac{3193}{350}}\)

\(=\dfrac{37}{266}+\dfrac{350}{3193}\)

\(=\dfrac{211241}{849338}\)

b)

\(N=\dfrac{3655}{11676}\Leftrightarrow\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{7+\dfrac{1}{a+\dfrac{1}{b}}}}}=\dfrac{3655}{11676}\)

\(\Leftrightarrow-\dfrac{36ab+36+5b}{115ab+115+16b}=\dfrac{3655}{11676}\)

dễ rồi lm tiếp nhé

Đó là hình thang vuông