+ )Cho \(x=99.\) Tính \(\text{A}=x^3+3x^2+3x\)
+)\(x+y=1.\text{Tính}\)\(B=3\cdot\left(x^2+y^2\right)-2\cdot\left(x^3-y^3\right)\)
+)\(x+y=101.\text{Tính}C=x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2012\)
tính giá trị biểu thức biết
\(\text{B=x^2-5-2xy+5y+y^2+2019 tại x=y+7}\)
\(\text{C=x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2019 tại x=101-y}\)
a)So sánh\(A=3^{32}-1\) và \(M=3\left(x^2+y^2\right)-2\left(x^3+y^3\right)\)
b) Cho x+y=101.Tính giá trị biểu thức
\(N=x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2012\)
b. \(N=x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2012\)\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(3x^2+6xy+3y^2\right)+\left(3x+3y\right)+2012\)
\(=\left(x+y\right)^3-3\left(x^2+2xy+y^2\right)+3\left(x+y\right)+2012\)
\(=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2012\) (*)
Thay x + y =101 vào biểu thức (*) ta được:
\(N=101^3-3.101^2+3.101+2012\)
= 1002013
Câu a ko hỉu đề!
Câu b:
Ta có: N = \(x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2012\)
= \(\left(x^3+3x^2y+3xy^2+y^3\right)-3\left(x^2+2xy+y^2\right)+3\left(x+y\right)+2012\)
= \(\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2012\)
= \(\left(x+y-1\right)^3+2013\)
Thay x + y = 101 vào N ta được:
N = 1003 + 2013 = 1002013
Phân tích đa thức thành nhân tử :
A = \(x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2015\)
B = \(x^2.\left(y-z\right)+y^2.\left(z-x\right)+z^2.\left(x-y\right)\)
B= x2(y-z)+y2(z-x)+z2(x-y)
= x2(y-z)+y2z-xy2+xz2-yz2
= x2(y-z)+yz(y-z)-x(y2-z2)
= x2(y-z)+yz(y-z)-x(y-z)(y+z)
= (y-z)(x2+yz -xy -xz)
= (y-z)[x(x-y)-z(x-y)]
= (y-z)(x-y)(x-z)
Bài 1: Phân tích đa thức thành nhân tử:
1) \(3x^3y^2-6xy\)
2) \(\left(x-2y\right).\left(x+3y\right)-2.\left(x-2y\right)\)
3) \(\left(3x-1\right).\left(x-2y\right)-5x.\left(2y-x\right)\)
4) \(x^2-y^2-6y-9\)
5) \(\left(3x-y\right)^2-4y^2\)
6) \(4x^2-9y^2-4x+1\)
8) \(x^2y-xy^2-2x+2y\)
9) \(x^2-y^2-2x+2y\)
Bài 2: Tìm x:
1) \(\left(2x-1\right)^2-4.\left(2x-1\right)=0\)
2) \(9x^3-x=0\)
3) \(\left(3-2x\right)^2-2.\left(2x-3\right)=0\)
4) \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
1) Cho x + y = 5. Tính:
\(P=3x^2-2x+3y^2-2y+6xy-100\\ Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
P = 3x2 - 2x + 3y2 - 2y + 6xy - 100
= (3x2 + 6xy + 3y2) - (2x + 2y) - 100
= 3(x2 + 2xy + y2) - 2(x + y) - 100
= 3(x + y)2 - 2.5 - 100
= 3. 52 -10 - 100
= 75 - 10 - 100 = -35
Q = x3 + y3 - 2x2 - 2y2 + 3xy(x + y) - 4xy + 3(x+y) +10
= x3 + y3 - 2x2 - 2y2 + 3x2y + 3xy2 - 4xy + 3.5 + 10
= (x3 + 3x2y + 3xy2 + y3) - (2x2 + 4xy + 2y2) + 15 + 10
= (x + y)3 - 2(x2 + 2xy + y2) + 25
= 53 - 2(x + y)2 +25
= 125 - 2. 52 + 25
= 125 - 50 + 25 = 100
1.Với giá trị nào của biến thì giá trị của biểu thức bằng 0
\(\frac{x+1}{7};\frac{3x+3}{5};\frac{3x\left(x-5\right)}{x-7};\frac{2x\left(x+1\right)}{3x+4}\)
2.Tính giá trị của các biểu thức sau:
\(A=\frac{a^2\left(a^2+b^2\right)\left(a^{\text{4}}+b^{\text{4 }}\right)\left(a^8+b^8\right)\left(a^2-3b\right)}{\left(a^{10}+b^{10}\right)}\)tại a=6;b=12
\(B=3xy\left(x+y\right)+2x^3y+2x^2y^2+5\)tại x+y=0
\(C=2x+2y+3xy\left(x+y\right)+5\left(x^3y^2+x^2y^3\right)+4\)tại x+y=0
Tính giá trị của biểu thức:
B=\(\text{x}^{\text{3}}+3\text{x}^{\text{2}}+3\text{x}^{\text{2}}y+3x\text{y}^{\text{2}}+\text{y}^{\text{3}}+3\text{y}^{\text{2}}+6xy+3x+3y+2019 \) Biết x+y=101
\(B=x^3+3x^2+3x^2y+3xy^2+y^3+3y^2+6xy+3x+3y+2019\)
\(=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2019\)
\(=\left[\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+1\right]+2018\)
\(=\left(x+y-1\right)^3+2018\)
Mà \(x+y=101\)
\(B=\left(101-1\right)^3+2018=1002018\)
cho x+y=5.tính giá tị biểu thức
P= \(^{ }3x^{^{ }2}-2x+3y^{^{ }2}-2y+6xy+2018\)
Q=\(x^3+y^{^{ }3}-2x-2y^{^{ }2}+3xy\left(x+y\right)-4xy-3x+3y\)
P = 3x2 - 2x + 3y2 - 2y + 6xy +2018
P = 3(x2 + y2 + 2xy) - 2(x + y) + 2018
P = 3[(x + y)2 - 2xy + 2xy] -2.5 + 2018
P = 3[ 52 +0] - 10 + 2018
P = 3.25 + 2008
P = 75 + 2008
P = 2083
1. Cho x=99. Tính giá trị biểu thức: x2+3x2+3x
2. Cho x+y=101. Tính giá trị biểu thức:
x^3 - 3x^2 + 3x^2y + 3xy^2 + y^2 - 3y^2 - 6xy + 3x + 3y + 2012
Làm giúp mình với ạ
1; \(x^2\) + 3\(x^2\) + 3\(x\) = 4\(x^2\) + 3\(x\) (1)
Thay \(x=99\) vào (1) ta có:
4.992 + 3.99 = 4.9801 + 297 = 39204 + 297 = 39501