Giải thích nữa nha

\(A=1+2^{3^{2012}}\)

\(\Rightarrow A=1+2^{6036}\)

\(1\equiv1\left(mod3\right)\)

\(2\equiv2\left(mod3\right)\)

\(\Rightarrow2^{6036}\equiv2\left(mod3\right)\)

\(\Rightarrow1+2^{6036}\equiv3\equiv0\left(mod3\right)\)

Vậy A là hợp số

a) \(ab-ac-b^2+2bc-c^2\)

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

\(=a\left(b-c\right)-\left(b-c\right)^2\)

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

b) \(x^6+8=\left(x^2\right)^3+2^3\)

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

c) \(64x^3-8=\left(4x\right)^3-2^3\)

\(=\left(4x-2\right)\left(16x^2+8x+4\right)\)

\(=8\left(2x-1\right)\left(4x^2+2x+1\right)\)

d) \(x^3-2x^2+4x-8\)

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

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

a) (2x - 1)(3x + 5) - 2(-4x + 1)² = 6x² + 10x - 3x - 5 - 2(16x² - 8x + 1) = 6x² - 3x - 5 - 32x² + 16x - 2 = -26x² + 13x - 7

b) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\frac{x+4}{x}\)

c) \(\frac{2x-9}{x^2-5x+6}+\frac{2x+1}{x-3}+\frac{x+3}{2-x}\)

= \(\frac{2x-9}{x^2-2x-3x+6}+\frac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)

= \(\frac{2x-9+2x^2-3x-2-x^2+9}{\left(x-3\right)\left(x-2\right)}\)

= \(\frac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}\)

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

= \(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\frac{x+1}{x-3}\)

d) (x - 1)³ - (x + 1)³ + 6(x + 1)(x - 1)

= (x - 1 - x - 1)[(x - 1)² + (x - 1)(x + 1) + (x + 1)²] + 6(x² - 1)

= -2(x² - 2x + 1  + x² - 1 + x² + 2x + 1) + 6x² - 6

= -2(3x² + 1) + 6x² - 6

= -6x² - 2 + 6x²  - 6

= -8

e) (2x + 7)² - (4x + 14)(2x - 8) + (8 - 2x)²

= (2x + 7)² - 2(2x + 7)(2x - 8) + (2x - 8)²

= (2x + 7 - 2x + 8)²

= 15² = 225

a) Ta có: x⁴ - x³ + 2x² - x + 1 = 0

=> (x⁴ + 2x² + 1) - x(x² + 1) = 0

=> (x² + 1)² - x(x² + 1) = 0

=> (x² + 1)(x² - x + 1) = 0

=> (x² + 1)[(x² - x + 1/4) + 3/4] = 0

=> (x²+  1 )[(x - 1/2)² + 3/4] = 0

=> pt vô nghiệm (vì x² + 1 > 0; (x - 1/2)² + 3/4 > 0)

b) Ta có: x³ + 2x² - 7x + 4 = 0

=> (x³ - x) + (2x² - 6x + 4) = 0

=> x(x² - 1) + 2(x² - 3x + 2) = 0

=> x(x - 1)(x + 1) + 2(x² - 2x - x + 2) = 0

=> x(x - 1)(x + 1) + 2(x - 2)(x - 1) = 0

=> (x - 1)(x² + x + 2x - 4) = 0

=> (x - 1)(x² + 3x - 4) = 0

=> (x - 1)(x²  + 4x - x - 4) = 0

=> (x - 1)(x + 4)(x - 1) = 0

=> (x - 1)²(x + 4) = 0

=> \(\orbr{\begin{cases}x-1=0\\x+4=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

a) \(x^4-x^3+2x^2-x+1=0\)

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

\(\Leftrightarrow\left(x^2+1\right)^2-x\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x^2+1-x\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left[\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}\right]=0\)

\(\Leftrightarrow\left(x^2+1\right)\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]=0\)

Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\end{cases}}\)

\(\Rightarrow\)Phương trình vô nghiệm

Vậy không có giá trị x thỏa mãn đề bài
 

b) \(x^3+2x^2-7x+4=0\)

\(\Leftrightarrow\left(x^3-x\right)+\left(2x^2-6x+4\right)=0\)

\(\Leftrightarrow x\left(x^2-1\right)+2\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+2\left(x^2-x-2x+2\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+2\left[x\left(x-1\right)-2\left(x-1\right)\right]=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+2\left(x-2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x+1\right)+2\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2+x+2x-4\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2+3x-4\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2+4x-x-4\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x+4\right)-\left(x+4\right)\right]=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-4\end{cases}}}\)

Vậy x=1; x=-4

Tham khảo: Câu hỏi của Bach Linh - Toán lớp 7 - Học toán với OnlineMath

link:https://olm.vn/hoi-dap/detail/64523861777.html

Lớp 7A nhận S là : 

\(\frac{300.15}{100}=45\left(m^2\right)\)

Lớp 7B nhận S là 

\(\frac{300-45}{5}=51\left(m^2\right)\)

Vậy suy ra 3 lớp còn lại nhận số S là : \(300-45-51=204\left(m^2\right)\)

Ta có : \(\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{4}}=\frac{c}{\frac{5}{16}}\)và \(a+b+c=204\)

== phần tiếp theo là toi ko chắc okey , ko bt có ADTC dãy tỉ số bằng nhau ko nha -.- 

ADTC dãy tỉ số bằng nhau 

\(\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{4}}=\frac{c}{\frac{5}{16}}=\frac{a+b+c}{\frac{1}{2}+\frac{1}{4}+\frac{5}{16}}=\frac{204}{\frac{17}{16}}=192\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{\frac{1}{2}}=192\\\frac{b}{\frac{1}{4}}=192\\\frac{c}{\frac{5}{16}}=192\end{cases}\Rightarrow\hept{\begin{cases}a=96\\b=48\\c=60\end{cases}}}\)

Tự KL nha ! 

1. \(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{x^2-1}\)

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

= \(\frac{-x-1+x-1+2}{\left(x-1\right)\left(x+1\right)}=0\)

c) \(\left(\frac{x^2-16}{x^2+8x+16}+\frac{6}{x+4}\right)\cdot\frac{2x}{x+2}\)

= \(\left(\frac{x^2-16}{\left(x+4\right)^2}+\frac{6\left(x+4\right)}{\left(x+4\right)^2}\right)\cdot\frac{2x}{x+2}\)

= \(\left(\frac{x^2-16+6x+24}{\left(x+4\right)^2}\right)\cdot\frac{2x}{x+2}\)

= \(\frac{x^2+6x+8}{\left(x+4\right)^2}\cdot\frac{2x}{x-2}\)

= \(\frac{x^2+4x+2x+8}{\left(x+4\right)^2}\cdot\frac{2x}{x+2}\)

= \(\frac{\left(x+4\right)\left(x+2\right)}{\left(x+4\right)^2}\cdot\frac{2x}{x+2}=\frac{2x}{x+4}\)

c/\(x^2-2x=2y-xy\)

\(x^2-2x+xy-2y=0\)

\(x\left(x-2\right)+y\left(x-2\right)=0\)

\(\left(x+y\right)\left(x-2\right)=0\)

d/\(x^2+4xy-16+4y^2\)

\(=\left(x^2+4xy+4y^2\right)-16\)

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

\(=\left(x+2y-4\right)\left(x+2y+4\right)\)

b/\(2x^2+7x-15\)

\(=2x^2+10x-3x-15\)

\(=\left(2x^2+10x\right)-\left(3x+15\right)\)

\(=2x\left(x+5\right)-3\left(x+5\right)\)

\(=\left(2x-3\right)\left(x+5\right)\)

Thêm một đk: a, b, c là số nguyên

Có: \(2x^3+15x^2+22x-15\)

\(=\left(2x^3-x^2\right)+\left(16x^2-8x\right)+\left(30x-15\right)\)

\(=x^2\left(2x-1\right)+8x\left(2x-1\right)+15\left(2x-1\right)\)

\(=\left(2x-1\right)\left(x^2+8x+15\right)\)

= \(\left(2x-1\right)\left[\left(x^2+3x\right)+\left(5x+15\right)\right]\)

\(=\left(2x-1\right)\left(x+3\right)\left(x+5\right)\)

Theo bài ra :  \(2x^3+15x^2+22x-15=\left(2x-a\right)\left(x+b\right)\left(x+c\right)\)

=> a + b + c  = 1 + 3 + 5 = 9.