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a)\(\frac{x^2+4}{x^2}+\frac{4}{x+1}\left(\frac{1}{x}+1\right)\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x+1}.\frac{x+1}{x}\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x}\)
\(=\frac{x^2+4x+4}{x^2}\)
\(\left(\frac{x+2}{x}\right)^2\)
=>phép chia = 1 với mọi x # 0 và x#-1
b)Cm tương tự
\(a,\frac{x^2+2.x.5+5^2+x^2-2.x.5+5^2}{x^2+25}\)
\(=\frac{2\left(x^2+25\right)}{x^2+25}=2\)
Vậy giá trị biểu thức không phụ thuộc vào x
\(b,\frac{4x^2+20x+25+25x^2-20x+4}{x^2+1}\)
\(=\frac{29\left(x^2+1\right)}{x^2+1}=29\)
Vậy gt biểu thức không phụ thuộc vào x
a) \(\frac{\left(x+5\right)^2+\left(x-5\right)^2}{x^2+25}\)
\(=\frac{x^2+10x+25+x^2-10x+25}{x^2+25}\)
\(=\frac{2\left(x^2+25\right)}{x^2+25}=2\)
\(\Rightarrow\)đpcm
b) \(\frac{\left(2x+5\right)^2+\left(5x-2\right)^2}{x^2+1}\)
\(=\frac{4x^2+20x+25+25x^2-20x+4}{x^2+1}\)
\(=\frac{29\left(x^2+1\right)}{x^2+1}=29\)
\(\Rightarrow\)đpcm
Với điều kiện xy\(\ne\)0;+ -3/2 y;x\(\ne\)-y các phân thức có nghĩa. Ta có
\(\frac{5x\left(2x-3y\right)^2}{3y\left(4x^2-9y^2\right)}:\frac{\left(2x^2+2xy\right)\left(2x-3y\right)}{2x^2y+5xy^2+3y^3}\)\(=\)\(\frac{5x\left(2x-3y\right)^2.y\left(2x^2+5xy+3y^2\right)}{3y\left(4x^2-9y^2\right).2x\left(x+y\right).\left(2x-3y\right)}\)
\(=\)\(\frac{10xy\left(2x-3y\right)^2.\left(2x^2+2xy+3xy+3y^2\right)}{6xy\left(2x-3y\right).\left(2x+3y\right)\left(x+y\right)\left(2x-3y\right)}\)\(=\)\(\frac{10xy\left(2x-3y\right)^2\left(x+y\right).\left(2x+3y\right)}{6xy\left(2x-3y\right)^2.\left(2x+3y\right).\left(x+y\right)}\)
\(=\)\(\frac{5}{3}\)
ĐK \(\hept{\begin{cases}xy\ne0\\2x-3y\ne0,2x+3y\ne0\\x\ne-y\end{cases}}\)
\(=\frac{5x\left(2x-3y\right)^2}{3y\left(2x+3y\right)\left(2x-3y\right)}:\frac{2x\left(x+y\right)\left(2x-3y\right)}{xy\left(2x+3y\right)+y^2\left(2x+3y\right)}\)
\(=\frac{5x\left(2x-3y\right)}{3y\left(2x+3y\right)}:\frac{2x\left(x+y\right)\left(2x-3y\right)}{\left(2x+3y\right)\left(xy+y^2\right)}\)
\(=\frac{5x\left(2x-3y\right)}{3y\left(2x+3y\right)}.\frac{y\left(x+y\right)\left(2x+3y\right)}{2x\left(x+y\right)\left(2x-3y\right)}=\frac{5}{6}\)
Vậy giá trị của biểu thức không phụ thuộc vào biến
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
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
= x2
Bài 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
a) Phân thức M xác định khi và chỉ khi :
+) \(2x-2\ne0\Leftrightarrow x\ne1\)
+) \(2x+2\ne0\Leftrightarrow x\ne-1\)
+) \(1-\frac{x-3}{x+1}\ne0\)
\(\Leftrightarrow x-3\ne x+1\)
\(\Leftrightarrow0x\ne4\left(\text{luôn đúng}\right)\)
Vậy \(x\ne\left\{1;-1\right\}\)
b) \(M=\left(\frac{x-2}{2x-2}-\frac{x+3}{2x+2}+\frac{3}{2x-2}\right):\left(1-\frac{x-3}{x+1}\right)\)
\(M=\left(\frac{\left(x-2\right)\left(2x+2\right)}{\left(2x-2\right)\left(2x+2\right)}-\frac{\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}+\frac{3\left(2x+2\right)}{\left(2x-2\right)\left(2x+2\right)}\right):\left(\frac{x+1-x+3}{x+1}\right)\)
\(M=\left(\frac{2x^2-2x-4-2x^2-4x+6+6x+6}{\left(2x-2\right)\left(2x+2\right)}\right):\left(\frac{4}{x+1}\right)\)
\(M=\frac{8}{2\left(x-1\right)2\left(x+1\right)}\cdot\frac{x+1}{4}\)
\(M=\frac{8\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)\cdot4}\)
\(M=\frac{8\left(x+1\right)}{8\left(x+1\right)\left(x-1\right)}\)
\(M=\frac{1}{x-1}\)
\(M=\left(\frac{x-2}{2x-2}-\frac{x+3}{2x+2}+\frac{3}{2x-2}\right):\left(1-\frac{x-3}{x+1}\right)\)
\(=\left(\frac{x+1}{2x-2}-\frac{x+3}{2x+2}\right):\left(\frac{4}{x+1}\right)=\left[\frac{\left(x+1\right)\left(2x+2\right)-\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]:\left(\frac{4}{x+1}\right)\)
\(=\left[\frac{2x^2+4x+2-2x^2+2x+6-6x+6}{4x^2-4}\right]:\left(\frac{4}{x+1}\right)\)
\(=\left[\frac{6x+8-6x+6}{4x^2-4}\right]:\left(\frac{4}{x+1}\right)\)
\(=\frac{14}{4x^2-4}:\left(\frac{4}{x+1}\right)=\frac{14x+14}{16x^2-16}=\frac{7x+7}{8x^2-8}\)
\(P=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{2x-5}{x^2+5x}+\frac{x}{5-x}\)
\(P=\left[\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right]:\frac{2x-5}{x\left(x+5\right)}+\frac{x}{5-x}\)
\(P=\frac{x^2-\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)x}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)
\(P=\frac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)
\(P=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)
\(P=\frac{5\left(2x-5\right).x\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)\left(2x-5\right)}+\frac{x}{5-x}\)
\(P=\frac{5}{x-5}+\frac{x}{5-x}\)
\(P=\frac{5}{x-5}-\frac{x}{x-5}\)
\(P=\frac{5-x}{x-5}\)
\(P=\frac{-\left(x-5\right)}{x-5}\)
\(P=-1\)
=> Giá trị của biểu thức P không phụ thuộc vào biến
đpcm