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`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`
`b)` Với `x ne -1;x ne -5` có:
`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`
`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`
`A=[x^2-3x-4]/[(x+1)(x+5)]`
`A=[(x+1)(x-4)]/[(x+1)(x+5)]`
`A=[x-4]/[x+5]`
`c)` Với `x ne -5; x ne -1; x ne 4` có:
`P=A.B=[x-4]/[x+5].[-10]/[x-4]`
`=[-10]/[x+5]`
Để `P` nguyên `<=>[-10]/[x+5] in ZZ`
`=>x+5 in Ư_{-10}`
Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`
`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)
d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-1}{x-2}\)
b: Khi x=1/2 thì \(B=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{2}{3}\)
Khi x=-1/2 thì B=2/5
c: Để B nguyên thì \(x-2\in\left\{1;-1\right\}\)
hay \(x\in\left\{3;1\right\}\)
a, đk : x khác -2 ; 2
\(B=\left(\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{1}{2-x}\)
b, Ta có \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{1}{2}\)
Với x = 1/2 ta được \(B=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)
Với x = -1/2 ta được \(B=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{2}{5}\)
c, \(\dfrac{1}{2-x}\Rightarrow2-x\inƯ\left(1\right)=\left\{\pm1\right\}\)
| 2-x | 1 | -1 |
| x | 1 | 3 |
a; A = (7\(x\) + 5)2 + (3\(x-5\))2 - (10 - 6\(x\)).(5 + 7\(x\))
A = 49\(x^2\) + 70\(x\) + 25 + 9\(x^2\) - 30\(x\) + 25 - 50 - 70\(x\) + 30\(x\) + 42\(x^2\)
A = (49\(x^2\) + 9\(x^2\) + 42\(x^2\)) + (70\(x-70x\)) - (30\(x\) - 30\(x\)) + (25+25-50)
A = 100\(x^2\) + 0 + 0 + (50 - 50)
A = 100\(x^2\) + 0 + 0 + 0
A = 100\(x^2\)
Thay \(x=-2\) vào A = 100\(x^2\) ta có:
A = 100.(-2)2
A = 100.4
A = 400.
a )
\(A=x\left(x^3+y\right)-x^2\left(x^2-y\right)-x^2\left(y-1\right)\)
\(\Rightarrow A=x^4+xy-x^4+x^2y-x^2y+x^2\)
\(\Rightarrow A=x^2+xy=x\left(x+y\right)\)
Thay \(x=-10;y=5\)vào A , ta được :
\(A=-10\left(-10+5\right)\)
\(=-10.-5=50\)
Vậy \(A=50\)
a) A = x(x3 + y) - x2(x2 - y) - x2(y - 1)
= x4 + xy - x4 + x2y - x2y + x2
= xy + x2
Thay x = –10 và y = 5 vào (1), ta được:
A = -10.5 + (-10)2 = -50 + 100 = 50
Vậy giá trị của biểu thức A tại x = –10 và y = 5 là 50.
b)Ta có: 5x3 – 3x2 + 10x – 6 = (5x3 + 10x )+ ( -3x2– 6)
= 5x(x2 + 2) – 3(x2 + 2) = (x2 + 2)(5x – 3)
Vậy (x2 + 2)(5x – 3) = 0 ⇒ 5x – 3 = 0 (vì x2 + 2 ≥ 0, với mọi x)
⇒x = 3/5
c)Ta có: x2 + y2 – 2x + 4y + 5 = (x2 – 2x + 1) + (y2 + 4y + 4)
= (x – 1)2 + (y + 2)2
Vậy (x – 1)2 + (y + 2)2 = 0 ⇒ x – 1 = 0 hay y + 2 = 0
⇒ x = 1 hoặc y = -2
\(a,A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{3+x}\right)\\ =\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{x^2-9}\right):\left(\dfrac{2\left(3+x\right)}{3+x}-\dfrac{x+5}{3+x}\right)\\ =\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{2\left(3+x\right)-\left(x+5\right)}{3+x}\\ =\left(\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{6+2x-x-5}{3+x}\)
\(=\dfrac{x^2-3x-\left(2x+6\right)-\left(x^2-1\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{3+x}\\ =\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3+x}{x+1}\\ =\dfrac{-5x-5}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3+x}{x+1}\\ =\dfrac{-5\left(x+1\right).\left(3+x\right)}{\left(x-3\right)\left(x+3\right).\left(x+1\right)}\\ =\dfrac{-5}{x-3}\)
\(b,A=x^2-x-2=0\\ \Leftrightarrow x^2+x-2x-2=0\\ \Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
\(c,\dfrac{-5}{x-3}=\dfrac{1}{2}\\ \Leftrightarrow-10=x-3\\ \Leftrightarrow-x+3=10\\ \Leftrightarrow-x=7\\ \Leftrightarrow x=7\)
Để `A=1/2` thì `x=7`
mình biết câu b rồi nhưng câu a thì chưa!
b) x^3(x+y)-x^2(x^2+xy)-x(x-y)
=x^4+x^3y-x^4-x^3y-x^2+xy
=-x^2+xy tại x=10,y=-5 ta có;
=-10^2+10(-5)
= 50
a: =>(x+10)(x-1)=0
=>x=-10 hoặc x=1
b: \(A=x^3-1-\left(x+5\right)\left(x^2-3\right)-5x^2-10x-5\)
\(=x^3-5x^2-10x-6-x^3+3x-5x^2+15\)
=-7x+9
=110/13