\(a,=\frac{2cos^2\alpha-cos^2\alpha-sin^2\alpha}{sin\alpha+cos\alpha}\\ =\frac{cos^2\alpha-sin^2\alpha}{sin\alpha+cos\alpha}\\ =cos\alpha-sin\alpha\)

\(b,sin25=cos65;cos70=sin20;Khiđó:B=1\)

Ta có sin25°=cos65°

         cos70°=20sin°

=> sịn25°+cos70°/sin20°+cos65°=cos65°+sin20°/sin20°+cos65°=1

Bài 1: 

\(=\left(\sin20^0-\cos70^0\right)+\left(-\tan40^0+\cot50^0\right)=0+0=0\)

Bài 2: 

\(\cos a=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}\)

\(A=2\cdot\sin^2a+3\cdot\cos^2a=2\cdot\dfrac{4}{9}+3\cdot\dfrac{5}{9}=\dfrac{8+15}{9}=\dfrac{23}{9}\)

Ta có :

\(A=\frac{a^2+2a}{2a+10}+\frac{a-5}{a}+\frac{50-5a}{2a\left(a+5\right)}\)

\(A=\frac{a^2+2a}{2\left(a+5\right)}+\frac{a-5}{a}+\frac{50-5a}{2a\left(a+5\right)}\)

a) Giá trị của biểu thức A xác định 

\(\Leftrightarrow\hept{\begin{cases}a+5\ne0\\a\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}a\ne-5\\a\ne0\end{cases}}}\)

Vậy để giá trị của biểu thức A xác định \(\Leftrightarrow\hept{\begin{cases}a\ne-5\\a\ne0\end{cases}}\)

ĐKXĐ : \(\hept{\begin{cases}a\ne-5\\a\ne0\end{cases}}\)

b) Ta có :

\(A=\frac{a^2+2a}{2\left(a+5\right)}+\frac{a-5}{a}+\frac{50-5a}{2a\left(a+5\right)}\)

\(A=\frac{a\left(a^2+2a\right)+2\left(a+5\right)\left(a-5\right)+50-5a}{2a\left(a+5\right)}\)

\(A=\frac{a^3+2a^2+2\left(a^2-25\right)+50-5a}{2a\left(a+5\right)}\)

\(A=\frac{a^3+4a^2-50+50-5a}{2a\left(a+5\right)}\)

\(A=\frac{a\left(a^2+4a-5\right)}{2a\left(a+5\right)}\)

\(A=\frac{a^2+5a-a-5}{2\left(a+5\right)}\)

\(A=\frac{\left(a+5\right)\left(a-1\right)}{2\left(a+5\right)}=\frac{a-1}{2}\)

c) Thay a = -1 ( Thỏa mãn ĐKXĐ ) vào biểu thức A ta có :

\(A=\frac{-1-1}{2}=-1\)

Vậy tại a = -1 thì giá trị của biểu thức A là - 1

d) Cho A = 0 , ta có :

\(\frac{a-1}{2}=0\)

\(\Leftrightarrow a-1=0\Leftrightarrow a=1\)( Thỏa mãn ĐKXĐ )

Vậy a = 1 thì giá trị của biểu thức A = 0 .

\(a.ĐKXĐ:\)\(2a+10\ne0\)            \(a\ne-5\)

                 \(a\ne0\)               \(\Leftrightarrow\)\(a\ne0\)     \(\Leftrightarrow\)\(\hept{\begin{cases}a\ne0\\a\ne-5\end{cases}}\)

                 \(2a\left(a+5\right)\ne0\)        \(\hept{\begin{cases}a\ne0\\a\ne-5\end{cases}}\)

\(b.A=\frac{a\left(a+2\right)}{2\left(a+5\right)}+\frac{a-5}{a}+\frac{5\left(10-a\right)}{2a\left(a+5\right)}\)

     \(=\frac{a\left(a+2\right)a}{2a\left(a+5\right)}+\frac{\left(a-5\right)2\left(a+5\right)}{2a\left(a+5\right)}+\frac{5\left(10-a\right)}{2a\left(a+5\right)}\)

   \(=\frac{a^3+2a^2+\left(2a-10\right)\left(a+5\right)+5\left(10-a\right)}{2a\left(a+5\right)}\)   

   \(=\frac{a^3+2a^2+2a^2+10a-10a-50+50-5a}{2a\left(a+5\right)}\)

   \(=\frac{a^3+4a^2-5a}{2a\left(a+5\right)}\) 

   \(=\frac{a\left(a^2+4a-5\right)}{2a\left(a+5\right)}\)

   \(=\frac{a\left(a-1\right)\left(a+5\right)}{2a\left(a+5\right)}\)

   \(=\frac{a-1}{2}\)với \(x\ne0\)và \(x\ne-5\)

\(c.\)Thay \(a=-1\left(t/mđk\right)\Leftrightarrow\frac{a-1}{2}\Rightarrow\frac{-1-1}{2}\)

                                          \(=-1\left(t/mđk\right)\)

\(d.A=0\Leftrightarrow A=\frac{a-1}{2}=0\)

                    \(\Rightarrow a-1=2.0\)

                    \(\Rightarrow a-1=0\)

                    \(\Rightarrow a=1\left(t/mđk\right)\)

a) Để P xác định \(\Leftrightarrow\hept{\begin{cases}2a-2\ne0\\2-2a^2\ne0\\a+2\ne0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a^2\ne1\\a\ne-2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a\ne-1vâ\ne1\\a\ne-2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a\ne-1\\a\ne2\end{cases}}\)

b) \(P=\left(\frac{a+1}{2a-2}+\frac{1}{2-2a^2}\right).\frac{2a+2}{a+2}\)

\(=\left[\frac{a+1}{2\left(a-1\right)}+\frac{1}{2\left(1-a\right)\left(1+a\right)}\right].\frac{2\left(a+1\right)}{a+2}\)

\(=\left[\frac{\left(a+1\right)^2}{2\left(a-1\right)\left(a+1\right)}-\frac{1}{2\left(a-1\right)\left(1+a\right)}\right].\frac{2\left(a+1\right)}{a+2}\)

\(=\frac{\left(a+1\right)^2-1}{2\left(a-1\right)\left(a+1\right)}.\frac{2\left(a+1\right)}{a+2}\)

\(=\frac{a\left(a+2\right)}{\left(a-1\right)\left(a+2\right)}\)

\(=\frac{a}{a-1}\)

c) \(\left|a\right|=3\Leftrightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)

+) Với a=3 thỏa mãn \(\hept{\begin{cases}a\ne1\\a\ne-1\\a\ne2\end{cases}}\)nên thay a=3 vào P ta được:

( làm nốt)

TH kia tương tự

a) \(\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)

\(=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)

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

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

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

\(\Leftrightarrow a^2+\sqrt{a}.\left(a-\sqrt{a}+1\right)-2\sqrt{a}.\left(a-\sqrt{a}+1\right)=2\left(a-\sqrt{a}+1\right)\)

\(\Leftrightarrow a^2-2\sqrt{a}.a+2a-\sqrt{a}-2a=2a-2\sqrt{a}+2\)

\(\Leftrightarrow a^2-2\sqrt{a}.a+2a-\sqrt{a}-2a=-2\sqrt{a}+2\)

\(\Leftrightarrow-2\sqrt{a}.a+2a-\sqrt{a}-2a=-2\sqrt{a}+2-a^2\)

\(\Leftrightarrow-2\sqrt{a}.a-\sqrt{a}=-2\sqrt{a}+2-a^2\)

\(\Leftrightarrow-2a\sqrt{a}+\sqrt{a}=2-a^2\)

\(\Leftrightarrow\sqrt{a}.\left(2a+1\right)=2-a^2\)

\(\Leftrightarrow\left[\sqrt{a}.\left(2a+1\right)\right]^2=\left(2-a^2\right)^2\)

\(\Leftrightarrow4a^3-4a^2+a=4-4a^2+a^4\)

\(\Leftrightarrow\orbr{\begin{cases}a=4\left(\text{thỏa mãn}\right)\\a=1\left(\text{loại}\right)\end{cases}}\)

=> a = 4

Cách ngắn hơn :

\(đkxđ\Leftrightarrow x\ge0\)

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

\(=\frac{\sqrt{a}\left(\sqrt{a}^3+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)

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

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

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

\(b,A=2\Rightarrow a-\sqrt{a}=2\)

\(\Rightarrow a-\sqrt{a}-2=0\)

\(\Rightarrow a+\sqrt{a}-2\sqrt{a}-2=0\)

\(\Rightarrow\sqrt{a}\left(\sqrt{a}+1\right)-2\left(\sqrt{a}+1\right)=0\)

\(\Rightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=2\\\sqrt{a}=-1\end{cases}\Rightarrow\orbr{\begin{cases}a=4\\a\in\varnothing\end{cases}}}\)

\(\Rightarrow a=4\)

\(c,A=a-\sqrt{a}=\sqrt{a}^2-2.\sqrt{a}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)

\(=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\)

\(\Rightarrow A_{min}=-\frac{1}{4}\Leftrightarrow\left(\sqrt{a}-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\sqrt{a}=\frac{1}{2}\Rightarrow a=\frac{1}{4}\)

Vậy với \(a=\frac{1}{4}\)thì A có giá trị nhỏ nhất là \(-\frac{1}{4}\)

Làm màu quá :))

a, ĐKXĐ: \(X\ne0;X\ne\pm1\)

b,\(A=\frac{X\left(X^2+2X+1\right)}{X\left(X^2-1\right)}=\frac{X\left(X+1\right)^2}{X\left(X-1\right)\left(X+1\right)}=\frac{X+1}{X-1}\)

c,Ta có: \(A=\frac{X+1}{X-1}=2\Leftrightarrow2\left(X-1\right)=X+1\Leftrightarrow2X-2=X+1\Leftrightarrow X=3\)

a) \(ĐKXĐ:x\ne0;x\ne1\)

b) \(A=\frac{x^3+2x^2+x}{x^3-x}\)

\(A=\frac{x\left(x^2+2x+1\right)}{x\left(x^2-1\right)}\)

\(A=\frac{x\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)

\(A=\frac{x+1}{x-1}\)

vậy \(A=\frac{x+1}{x-1}\)

c) thay vào ta được \(\frac{x+1}{x-1}=2\)

\(\Rightarrow\left(x-1\right).2=x+1\)

\(\Rightarrow2x-2=x+1\)

\(\Rightarrow2x-x=1+2\)

\(\Rightarrow x=3\)

vậy \(x=3\)thì \(A=2\)