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a) Rút gọn :
\(ĐKXĐ:x\ne\pm5\)
Ta có : \(P=\left(\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right):\frac{2x-5}{x\left(x+5\right)}-\frac{2x}{5-x}\)
\(=\left(\frac{x^2-\left(x-5\right)\left(x-5\right)}{x\left(x-5\right)\left(x+5\right)}\right):\frac{\left(2x-5\right)\left(x-5\right)+2x^2\left(x+5\right)}{x\left(x+5\right)\left(x-5\right)}\)
\(=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\frac{x\left(x+5\right)\left(x-5\right)}{ }\)
Tui đang định làm tiếp đó, nhưng khẳng định đề này hơi sai sai ở vế bị chia. Bạn xem lại đc k ?
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) Đk: x > 0 và x khác +-1
Ta có: A = \(\left(\frac{x+1}{x}-\frac{1}{1-x}-\frac{x^2-2}{x^2-x}\right):\frac{x^2+x}{x^2-2x+1}\)
A = \(\left[\frac{\left(x-1\right)\left(x+1\right)+x-x^2+2}{x\left(x-1\right)}\right]:\frac{x\left(x+1\right)}{\left(x-1\right)^2}\)
A = \(\frac{x^2-1+x-x^2+2}{x\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{x\left(x+1\right)}\)
A = \(\frac{x+1}{x}\cdot\frac{x-1}{x\left(x+1\right)}=\frac{x-1}{x^2}\)
b) Ta có: A = \(\frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}=-\left(\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\frac{1}{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> 1/x - 1/2 = 0 <=> x = 2 (tm)
Vậy MaxA = 1/4 <=> x = 2
\(a,\)\(đkxđ\Leftrightarrow\)\(\hept{\begin{cases}x+3\ne0\\x-3\ne0\end{cases}}\)\(\Rightarrow x\ne\pm3\)
\(b,\)\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{x^2-9}\)
\(=\frac{5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{5x-15+3x+9-5x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x-9}{\left(x-3\right)\left(x+3\right)}=\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3}{x+3}\)
\(c,\)Tại x = 6, ta có :
\(B=\frac{3}{x+3}=\frac{3}{6+3}=\frac{3}{9}=\frac{1}{3}\)
Vậy tại x = 6 thì B = 3
\(d,\)Để \(B\in Z\Rightarrow\frac{3}{x+3}\in Z\Rightarrow x+3\inƯ_3\)
Mà \(Ư_3=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\)TH1 : \(x+3=1\Rightarrow x=-2\)
Th2: \(x+3=-1\Rightarrow x=-4\)
Th3 : \(x+3=3\Rightarrow x=0\)
TH4 \(x+3=-3\Rightarrow x=-6\)
Vậy để \(B\in Z\)thì \(x\in\left\{-6;-4;-2;0\right\}\)
a)Để B đc xác định thì :x+3 khác 0
x-3 khác 0
x^2-9 khác 0
=>x khác -3
x khác 3
b) Kết Qủa BT B là:3/x+3
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm5\end{cases}}\)
\(M=\left(\frac{x}{x+5}-\frac{5}{5-x}+\frac{10x}{x^2-25}\right)\cdot\left(1-\frac{5}{x}\right)\)
\(\Leftrightarrow M=\frac{x^2-5x+5x+25+10x}{\left(x+5\right)\left(x-5\right)}\cdot\frac{x-5}{x}\)
\(\Leftrightarrow M=\frac{\left(x^2+10x+25\right)\left(x-5\right)}{\left(x+5\right)\left(x-5\right)x}\)
\(\Leftrightarrow M=\frac{\left(x+5\right)^2}{x\left(x+5\right)}\)
\(\Leftrightarrow M=\frac{x+5}{x}\)
b) Để \(M\inℤ\)
\(\Leftrightarrow x+5⋮x\)
\(\Leftrightarrow5⋮x\)
\(\Leftrightarrow x\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Mà \(x\ne\pm5\)
\(\Leftrightarrow x\in\left\{1;-1\right\}\)
Vậy để \(M\inℤ\Leftrightarrow x\in\left\{1;-1\right\}\)
\(M=\left(\frac{x}{x+5}-\frac{5}{5-x}+\frac{10x}{x^2-25}\right)\cdot\left(1-\frac{5}{x}\right)\left(x\ne\pm5;x\ne0\right)\)
\(\Leftrightarrow M=\left(\frac{x}{x+5}+\frac{5}{x-5}+\frac{10x}{\left(x-5\right)\left(x+5\right)}\right)\cdot\frac{x-5}{x}\)
\(\Leftrightarrow M=\left(\frac{x^2-5x}{\left(x-5\right)\left(x+5\right)}+\frac{5x+25}{\left(x-5\right)\left(x+5\right)}+\frac{10x}{\left(x-5\right)\left(x+5\right)}\right)\cdot\frac{x-5}{x}\)
\(\Leftrightarrow M=\frac{x^2-5x+5x+25+10x}{\left(x-5\right)\left(x+5\right)}\cdot\frac{x-5}{x}\)
\(\Leftrightarrow M=\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}\cdot\frac{x-5}{x}\)
\(\Leftrightarrow M=\frac{\left(x+5\right)^2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)x}=\frac{x+5}{x}\)
b) M là số nguyên thì x+5 chia hết cho x
=> 5 chia hết cho x
x nguyên => x thuộc Ư (5)={-5;-1;1;5}
Vậy x={-5;-1;1;5} thì M là số nguyên
\(A=\left(a+2b-5+b\right)^2-2ab+34=\left(a+2b-5\right)^2+2b\left(a+2b-5\right)+b^2-2ab+34\)
\(A=\left(a+2b-5\right)^2+5b^2-10b+5+29\)
\(A=\left(a+2b-5\right)^2+5\left(b-1\right)^2+29\ge29\)
\(A_{min}=29\) khi \(\hept{\begin{cases}a=3\\b=1\end{cases}}\)
\(B=x+\frac{25}{x}-8\ge2\sqrt{x.\frac{25}{x}}-8=2\)
\(B_{min}=2\) khi \(x=5\)
\(C=\frac{x^2-15x+36}{x}=x+\frac{36}{x}-15\ge2\sqrt{x.\frac{36}{x}}-15=-3\)
\(C_{min}=-3\) khi \(x=6\)
Nếu không đọc được công thức của E thì các bạn nhấp vào câu hỏi nha!