Câu hỏi của Mr Black

Question

Bài 1) Rút gọn biểu thức$\left(x-y+z\right)^2+\left(z-y\right)^2+\left(x-y+z\right)\left(2y-2z\right)$Bài 2) Chứng minh giá trị của biểu thức không phụ thuộc vào biến\(\left(\frac{1}{x^2+x}+\frac{1}...

Nguyễn Minh Đăng · Accepted Answer

Bài 1:$\left(x-y+z\right)^2+\left(z-y\right)^2+\left(x-y+z\right)\left(2y-2z\right)$$=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2$$=\left(x-y+z+y-z\right)^2$$=x^2$Bài 2:đk: $x\ne\left\{0;-1;-2;-3;-4;-5\right\}$Xét BT trái ta có:$\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}$$=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}$$=\frac{1}{x}-\frac{1}{x+5}$$=\frac{5}{x\left(x+5\right)}=\frac{5}{x^2+5x}$GT của biểu thức lớn sẽ là: $\frac{5}{x^2+5x}\cdot\frac{x^2+5x}{5}=1$ không phụ thuộc vào biến=> đpcmCâu trả lời: Bài 1:$\left(x-y+z\right)^2+\left(z-y\right)^2+\left(x-y+z\right)\left(2y-2z\right)$$=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2$$=\left(x-y+z+y-z\right)^2$$=x^2$Bài 2:đk: $x\ne\left\{0;-1;-2;-3;-4;-5\right\}$Xét BT trái ta có:$\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}$$=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}$$=\frac{1}{x}-\frac{1}{x+5}$$=\frac{5}{x\left(x+5\right)}=\frac{5}{x^2+5x}$GT của biểu thức lớn sẽ là: $\frac{5}{x^2+5x}\cdot\frac{x^2+5x}{5}=1$ không phụ thuộc vào biến=> đpcm

l҉o҉n҉g҉ d҉z҉ · Answer

Bài 1.( x - y + z ) + ( z - y )2 + ( x - y + z )( 2y - 2z )= ( x - y + z ) - 2( x - y + z )( z - y ) + ( z - y )2= [ ( x - y + z ) - ( z - y ) ]2 = ( x - y + z - z + y )2= x2Bài 2. ĐKXĐ tự ghi nhé :))$\left(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\right)\times\left(\frac{x^2+5x}{5}\right)$$=\left(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)$$=\left(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)$$=\left(\frac{1}{x}-\frac{1}{x+5}\right)\times\frac{x\left(x+5\right)}{5}$$=\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{\left(x+5\right)}\right)\times\frac{x\left(x+5\right)}{5}$$=\frac{x+5-x}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}$$=\frac{5}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}=1$=> đpcmCâu trả lời: Bài 1.( x - y + z ) + ( z - y )2 + ( x - y + z )( 2y - 2z )= ( x - y + z ) - 2( x - y + z )( z - y ) + ( z - y )2= [ ( x - y + z ) - ( z - y ) ]2 = ( x - y + z - z + y )2= x2Bài 2. ĐKXĐ tự ghi nhé :))$\left(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\right)\times\left(\frac{x^2+5x}{5}\right)$$=\left(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)$$=\left(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)$$=\left(\frac{1}{x}-\frac{1}{x+5}\right)\times\frac{x\left(x+5\right)}{5}$$=\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{\left(x+5\right)}\right)\times\frac{x\left(x+5\right)}{5}$$=\frac{x+5-x}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}$$=\frac{5}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}=1$=> đpcm

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