a: \(=\dfrac{-\dfrac{1}{2}\left[cos\left(a+b+a-b\right)-cos\left(a+b-a+b\right)\right]}{cos^2b-cos^2a}\)

\(=\dfrac{-\dfrac{1}{2}\cdot\left[cos2a-cos2b\right]}{\dfrac{1-cos2b}{2}-\dfrac{1-cos2a}{2}}\)

\(=\dfrac{-\dfrac{1}{2}\cdot\left(cos2a-cos2b\right)}{\dfrac{1-cos2b-1+cos2a}{2}}=\dfrac{-\dfrac{1}{2}\cdot\left(cos2a-cos2b\right)}{\dfrac{1}{2}\cdot\left(cos2a-cos2b\right)}=-1\)

c: \(T=\dfrac{sina+sinb\cdot\left(cosa\cdot cosb-sina\cdot sinb\right)}{cosa-sinb\cdot\left(sina\cdot cosb+sinb\cdot cosa\right)}-tan\left(a+b\right)\)

\(=\dfrac{sina+sinb\cdot cosa\cdot cosb-sin^2b\cdot sina}{cosa-sinb\cdot sina\cdot cosb-sin^2b\cdot cosa}-tan\left(a+b\right)\)

\(=\dfrac{sina\left(1-sin^2b\right)+sinb\cdot cosa\cdot cosb}{cosa\left(1-sin^2b\right)-sinb\cdot sina\cdot cosb}\)-tan(a+b)

\(=\dfrac{sina\cdot cos^2b+sinb\cdot cosa\cdot cosb}{cosa\cdot cos^2b-sinb\cdot sina\cdot cosb}-tan\left(a+b\right)\)

\(=\dfrac{sina\cdot cosb+sinb\cdot cosa}{cosa\cdot cosb-sina\cdot sinb}-tan\left(a+b\right)\)

\(=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}-tan\left(a+b\right)=0\)

Với số to như 25764 bạn muốn kiểm tra có chia hết cho 8 không thì có thể lấy 3 chữ số cuối cùng. Nếu 3 chữ số cuối cùng tạo thành số chia hết cho 8 thì cả số đó chia hết cho 8 còn không thì không chia hết.

Trong TH này 764 không chia hết cho 8 nên 25764 cũng không chia hết cho 8.

Câu 2

\((1) MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O\\ (2) Cl_2 + H_2 \xrightarrow{as} 2HCl\\ (3) 3Cl_2 + 2Fe \xrightarrow{t^o} 2FeCl_3\\ (4) 2FeCl_3 + Fe \to 3FeCl_2\\ (5) 2NaOH + Cl_2 \to NaCl + NaClO + H_2O\)

\((1) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ (2) 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ (3) C + O_2 \xrightarrow{t^o} CO_2\\ (4) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ (5) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ (6) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ (7) Fe + H_2SO_4 \to FeSO_4 + H_2\\ (8) Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O\\ (9) 2Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O\\ (10) 2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O\)

4.

\(\lim\limits_{x\rightarrow8}f\left(x\right)=\lim\limits_{x\rightarrow8}\dfrac{\sqrt[3]{x}-2}{x-8}=\lim\limits_{x\rightarrow8}\dfrac{x-8}{\left(x-8\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}=\lim\limits_{x\rightarrow8}\dfrac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}\)

\(=\dfrac{1}{4+4+4}=\dfrac{1}{12}\)

\(f\left(8\right)=3.8-20=4\)

\(\Rightarrow\lim\limits_{x\rightarrow8}f\left(x\right)\ne f\left(8\right)\)

\(\Rightarrow\) Hàm gián đoạn tại \(x=8\)

5.

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt[]{1+2x}-1+1-\sqrt[3]{1+3x}}{x}=\lim\limits_{x\rightarrow0^+}\dfrac{\dfrac{2x}{\sqrt[]{1+2x}+1}-\dfrac{3x}{1+\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0^+}\left(\dfrac{2}{\sqrt[]{1+2x}+1}-\dfrac{3}{1+\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}\right)=\dfrac{2}{1+1}-\dfrac{3}{1+1+1}=0\)

\(f\left(0\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\left(3x^2-2x\right)=0\)

\(\Rightarrow\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=f\left(0\right)\)

\(\Rightarrow\) Hàm liên tục tại \(x=0\)

6.

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt[]{4x+1}-\sqrt[3]{6x+1}}{x^2}\)

\(=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt[]{4x+1}-\left(2x+1\right)+\left(2x+1-\sqrt[3]{6x+1}\right)}{x^2}\)

\(=\lim\limits_{x\rightarrow0^+}\dfrac{\dfrac{-x^2}{\sqrt[]{4x+1}+2x+1}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0^+}\left(\dfrac{-1}{\sqrt[]{4x+1}+2x+1}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}\right)\)

\(=\dfrac{-1}{1+1}+\dfrac{12}{1+1+1}=\dfrac{7}{2}\)

\(f\left(0\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\left(2-3x\right)=2\)

\(\Rightarrow\lim\limits_{x\rightarrow0^+}f\left(x\right)\ne\lim\limits_{x\rightarrow0^-}f\left(x\right)\)

\(\Rightarrow\) Hàm gián đoạn tại \(x=0\)

7.

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt[]{1+2x}-\left(x+1\right)+\left(x+1-\sqrt[3]{1+3x}\right)}{x^2}\)

\(=\lim\limits_{x\rightarrow0^+}\dfrac{\dfrac{-x^2}{\sqrt[]{1+2x}+x+1}+\dfrac{x^2\left(x+3\right)}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0^+}\left(\dfrac{-1}{\sqrt[]{1+2x}+x+1}+\dfrac{x+3}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}\right)\)

\(=\dfrac{-1}{1+1}+\dfrac{3}{1+1+1}=1\)

\(f\left(0\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\left(2x+3\right)=3\)

\(\Rightarrow\lim\limits_{x\rightarrow0^+}f\left(x\right)\ne\lim\limits_{x\rightarrow0^-}f\left(x\right)\)

\(\Rightarrow\) Hàm gián đoạn tại \(x=0\)

Lớp 7 học cái này rồi bạn

Cái này chỉ đơn giản là AB=0 thì A=0 hoặc B=0 thôi

\(7,x^4+x^3+x^2-1=x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)=\left(x^3+x-1\right)\left(x+1\right)\)

\(8,x^2y^2+1-x^2-y^2=\left(x^2y^2-y^2\right)-\left(x^2-1\right)\\ =y^2\left(x^2-1\right)-\left(x^2-1\right)=\left(x-1\right)\left(x+1\right)\left(y-1\right)\left(y+1\right)\)

\(10,x^4-x^2+2x-1=x^4-\left(x-1\right)^2=\left(x^2-x+1\right)\left(x^2+x-1\right)\\ 11,3a-3b+a^2-2ab+b^2=3\left(a-b\right)+\left(a-b\right)^2=\left(3+a-b\right)\left(a-b\right)\\ 12,a^2+2ab+b^2-2a-2b+1=\left(a+b\right)^2-2\left(a+b\right)+1=\left(a+b-1\right)^2\\ 13,a^2-b^2-4a+4b=\left(a-b\right)\left(a+b\right)-4\left(a-b\right)=\left(a+b-4\right)\left(a-b\right)\\ 14,a^3-b^3-3a+3b=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)=\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ 15,x^3+3x^2-3x-1=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=\left(x-1\right)\left(x^2+4x+1\right)\)

1)
=0,25y.(64x³+z³)
2)
=x²(x²-4x+4)
=x²(x-2)²
5)
=x²(x+1)-4(x+1)
=(x²-4)(x+1)
=(x-2)(x+2)(x+1)
6)
=x²(x-1)-(x-1)
=(x²-1)(x-1)
=(x-1)(x+1)(x-1)
=(x-1)²(x+1)