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1. ĐKXĐ: \(x\ne\pm1\)
2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)
\(=\dfrac{\left(x+1\right)^2-\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{x^2+2x+1-x^2+4x-3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x-3}{x-1}\)
3. Tại x = 5, A có giá trị là:
\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)
4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)
Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)
Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)
A xác định khi 5x-10 ≠0 <=> X ≠ 2b) A = x²-4x+4/5x-10= (x-2)²/5(x-2)= x-2/5c) x= -2018<=> A = -2018-2/5= -2020/5 = -404
Chúc bạn học tốt
a) ĐKXĐ: \(x\ne2\)
b) Ta có: \(A=\dfrac{x^2-4x+4}{5x-10}\)
\(=\dfrac{\left(x-2\right)^2}{5\left(x-2\right)}\)
\(=\dfrac{x-2}{5}\)
a. ĐKXĐ: \(x\ne\pm1\)
b. \(A=\left(x^2-1\right)\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left[\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right]\)
\(=\left(x-1\right)\left(x+1\right)\left[\dfrac{x+1-x+1-\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right]\)
\(=\left(x-1\right)\left(x+1\right)\left[\dfrac{-x^2+3}{\left(x-1\right)\left(x+1\right)}\right]\)
\(=\dfrac{\left(x-1\right)\left(x+1\right)\left(-x^2+3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=-x^2+3\)
c. Thay x = 3 vào A ta được:
\(-\left(3\right)^2+3=-6\)
Vậy: Giá trị của A tại x = 3 là -6
a) ĐKXĐ: \(x\ne1;x\ne-1.\)
b) \(A=\left(x^2-1\right).\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}-1\right).\)
\(=\left(x^2-1\right).\dfrac{x+1-x+1-x^2+1}{x^2-1}=-x^2+3.\)
c) Thay x = 3 (TMĐK) vào A: \(-3^2+3=-6.\)
a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{x-2}\)
Câu 2:
a: ĐKXĐ: \(x\notin\left\{0;2\right\}\)
b: Sửa đề: \(A=\left(\dfrac{2x-x^2}{2x^2+8}-\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\left(\dfrac{2}{x^2}-\dfrac{x-1}{x}\right)\)
\(=\left(\dfrac{2x-x^2}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{2-x\left(x-1\right)}{x^2}\)
\(=\left(\dfrac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{2-x^2+x}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x^3-2x^2-2x^2+4x+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3+4x}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)\left(x+1\right)}{2\left(x^2+4\right)\cdot x^2}=\dfrac{x+1}{2x}\)
c: Khi x=2024 thì \(A=\dfrac{2024+1}{2\cdot2024}=\dfrac{2025}{4048}\)
Câu 1:
a: \(25x^2\left(x-3y\right)-15\left(3y-x\right)\)
\(=25x^2\left(x-3y\right)+15\left(x-3y\right)\)
\(=\left(x-3y\right)\left(25x^2+15\right)\)
\(=\left(x-3y\right)\cdot5\cdot\left(5x^2+3\right)\)
b: \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=\left(x^4-x^2\right)-\left(4x^2-4\right)\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
a) ĐKXĐ: \(x\ne0;x\ne-2\)
b) \(S=\dfrac{\left(x+2\right)^2}{x}\cdot\left(1-\dfrac{x^2}{x+2}\right)-\dfrac{x^2+6x+4}{x}\)
\(=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{x+2-x^2}{x+2}-\dfrac{x^2+6x+4}{x}\)
\(=\dfrac{\left(x+2\right)\left(x+2-x^2\right)}{x}-\dfrac{x^2+6x+4}{x}\)
\(=\dfrac{x^2+2x-x^3+2x+4-2x^2-x^2-6x-4}{x}\)
\(=\dfrac{-x^3-2x^2-2x}{x}\)
\(=\dfrac{x\left(-x^2-2x-2\right)}{x}\)
\(=-x^2-2x-2\)
Với \(x=0\Rightarrow\) loại
Với \(x=1\), thay vào \(S\) ta được
\(S=-1^2-2\cdot1-2=-5\)
c) Có: \(S=-x^2-2x-2\)
\(=-\left(x^2+2x+2\right)\)
\(=-\left(x^2+2x+1\right)-1\)
\(=-\left(x+1\right)^2-1\)
Ta thấy: \(\left(x+1\right)^2\ge0\forall x\ne0;x\ne-2\)
\(\Rightarrow-\left(x+1\right)^2\le0\forall x\ne0;x\ne-2\)
\(\Rightarrow S=-\left(x+1\right)^2-1\le-1\forall x\ne0;x\ne-2\)
Dấu \("="\) xảy ra khi: \(x+1=0\Leftrightarrow x=-1\left(tmdk\right)\)
\(\text{#}\mathit{Toru}\)
a) a ≠ 0 , a ≠ − 5
b) Ta có A = a 3 + 4 a 2 − 5 a 2 a ( a + 5 ) = a ( a − 1 ) ( a + 5 ) 2 a ( a + 5 ) = a − 1 2
c) Thay a = -1 (TMĐK) vào a ta được A = -1
d) Ta có A = 0 Û a = 1 (TMĐK)
ĐK: \(x\ne\pm2\)
\(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right).\dfrac{x+2}{2}\)
\(=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right].\dfrac{x+2}{2}\)
\(=\dfrac{x-2x-2+x-2}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{2}\)
\(=\dfrac{2}{2-x}\)
a) đkxđ
\(\hept{\begin{cases}x+5\ne0\\2x^2+10\ne0\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne-5\\2x^2\ne-10\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne5\\x^2\ne-5\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne5\\x\ne\sqrt{5};\sqrt{-5}\\x\ne0\end{cases}}\)
Vậy .....
b) Rút gọn
\(\Rightarrow A=\frac{x^2+2x}{x+5}+\frac{50-5x}{2x\left(x+5\right)}+\frac{x-5}{x}\)
MTC: 2x(x+5)
\(\Rightarrow A=\frac{\left(x^2+2x\right).2x+50-5x+2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}\)
\(\Rightarrow A=\frac{2x^3+4x+50-5x+2.\left(x^2-25\right)}{2x\left(x+5\right)}\)
\(\Rightarrow A=\frac{2x^3+4x+50-5x+2x^2-50}{2x\left(x+5\right)}\)
\(\Rightarrow A=\frac{2x^3-x+2x^2}{2x\left(x+5\right)}\)
\(\Rightarrow A=\frac{x\left(2x^2-1+2x\right)}{2x\left(x+5\right)}\)
\(\Rightarrow A=\frac{2x^2-1+2x}{x\left(x+5\right)}\)
Vậy với ( ghi đkxđ) thì A=_____
c) Thay x=-4 vào A =_____ ta có:
\(A=\frac{2x^2-1+2x}{x\left(x+5\right)}=\frac{2.\left(-4\right)^2-1+2.\left(-4\right)}{4.\left(4+5\right)}=\frac{-1}{36}\)
Vậy với x=-4 thì A=-1/36
d)
Ta có để giá trị A=-3/2 thì
\(\frac{2x^2-1+2x}{x\left(x+5\right)}=-\frac{3}{2}\)
\(\Rightarrow2.\left(2x^2-1+2x\right)=-3x\left(x+5\right)\)
\(\Rightarrow4x^2-2+4x=-3x^2-15x\)
\(\Rightarrow4x^2-2+4x+3x^2-15x=0\)
\(\Rightarrow7x^2-2-11x=0\)
Ta có: vì 7x^2 > 0
=> 7x^2-2-11x > 0 ( k/tm)
=> o có gt nào của x để A=-3/2
( chờ xíu mình check b lại uwu)