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\(5x^2-\left(2x+1\right)\left(x-2\right)-x\left(3x+3\right)+7\)
\(=5x^2-2x^2+4x-x+2-3x^2-3x+7\)
=9
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
Sửa đề: \(5x^2+5y^2+8xy-2x+2y+2=0\)
=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)
=>\(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
=>\(\left\{{}\begin{matrix}2x+2y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(M=\left(x-y\right)^{2023}-\left(x-2\right)^{2024}+\left(y+1\right)^{2023}\)
\(=\left(1+1\right)^{2023}-\left(1-2\right)^{2024}+\left(-1+1\right)^{2023}\)
\(=2^{2023}-1\)
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2=x.5x^2+x.\left(-3x\right)+x.2+\left(-1\right).5x^2+\left(-1\right)\left(-3x\right)+\left(-1\right).2=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
b: \(\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)
\(=x^2-2x+1\)
\(=\left(x-1\right)^2\)
c: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=5x^3+14x^2+12x+8\)
Ta có P = 5 x 2 − [ 4 x 2 − 3 x ( x − 2 ) ]
= 5 x 2 – (4 x 2 – 3 x 2 + 6x) = 5 x 2 – ( x 2 + 6x)
= 5 x 2 – x 2 – 6x = 4 x 2 – 6x
Thay x = − 3 2 vào biểu thức P = 4 x 2 – 6x ta được
P = 4. ( − 3 2 ) 2 − 6. ( − 3 2 ) = 4. 9 4 + 18 2 = 18
Vậy P = 4 x 2 – 6x. Với x = − 3 2 thì P = 18
Đáp án cần chọn là: A
1/
a)5x – 20y=5(x-4y)
b) 5x.(x – 1) – 3x(x – 1)=2x(x-1)
c) x.(x+y) – 5x – 5y=c) x.(x+y) – 5(x+y)=(x-5)(x+y)
2/
a)x2 + xy + x = x(x+y+1)=77.(77+22+1)=77.100=7700
b) x . ( x – y ) + y . ( y – x )=(x-y)(x-y)=(x-y)2=(53-3)2=2500
3/
a) X + 5x2 = 0
⇒x(x+5)=0
⇒hoặc x=0
x+5=0⇒x=-5
b)x + 1 = ( x + 1 )2
⇒(x + 1)-( x + 1 )2 =0
⇒x(x+1)=0
⇒ hoặc x=0
hoặc x+1=0⇒x=-1
\(a,=5\left(x^2+2xy+y^2\right)-10y^2+5=5\left(x+y\right)^2-10y^2+5\\ =5\left(1+2\right)^2-10\cdot4+5=45-40+5=10\\ b,=7\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(7-x+y\right)\\ =\left(2-2\right)\left(7-2+2\right)=0\)
b: \(=7\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(7-x+y\right)=0\)
A=\(3*(5x^2-2)-5x^2(7+3x)-2,5(2-14x^2)\) với \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow A=15x^2-6-35x^2-15x^3-5+35x^2=-15x^3+15x^2-11\)
Với \(x=\dfrac{1}{2}\Rightarrow A=-15\cdot\left(\dfrac{1}{2}\right)^3+15\cdot\left(\dfrac{1}{2}\right)^2-11=-\dfrac{73}{8}\)
\(Với\) \(x=-\dfrac{1}{2}\Rightarrow A=-15\cdot\left(-\dfrac{1}{2}\right)^3+15\cdot\left(-\dfrac{1}{2}\right)^2-11=-\dfrac{43}{8}\)