Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐKXĐ : \(\left\{{}\begin{matrix}4x^2-1\ne0\\8x^3+1\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm\dfrac{1}{2}\)
\(P=\dfrac{2x^5-x^4-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)
\(=\dfrac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(=\dfrac{x^4-1}{2x+1}+\dfrac{2}{2x+1}=\dfrac{x^4+1}{2x+1}\)
7: Ta có: \(\left(3x+4\right)\left(2x-1\right)+6x\left(1-x\right)=0\)
\(\Leftrightarrow6x^2-3x+8x-4+6x-6x^2=0\)
\(\Leftrightarrow11x=4\)
hay \(x=\dfrac{4}{11}\)
8: Ta có: \(2x\left(x^2-1\right)+x\left(-2x^2-3x+1\right)=-x-27\)
\(\Leftrightarrow2x^3-2x-2x^3-3x^2+x+x+27=0\)
\(\Leftrightarrow x^2=9\)
hay \(x\in\left\{3;-3\right\}\)
\(P=\dfrac{-x^4+2x^3-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)
\(=\dfrac{\left(1-x^2\right)\left(1+x^2\right)+2x\left(x^2-1\right)}{4x^2-1}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(=\dfrac{\left(1-x^2\right)\left(1+x^2-2x\right)}{4x^2-1}+\dfrac{2}{2x+1}\)
\(=\dfrac{\left(1-x^2\right)\left(x^2-2x+1\right)+4x-2}{4x^2-1}\)
a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)
\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)
\(=6x^2-3x+\dfrac{5}{2}\)
b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)
\(=3x-y-y-x+2x^2-2x\)
\(=2x^2-2y\)
câu 1:
a,x2+2x-4z2+1
=x2+2x.1+12-(2z)2
=(x+1)2-(2z)2
=(x+1-2z)(x+1+2z)
<>?/[;b[]rwel;u];53pjkjnlgkljtreylkeuro;uwqr[i5uiwehhwwejokejoiyufljukneghnmknbfvhdbg.elkgiwr;iewqirluoyeiwhtgo
Đặt bthuc = A nhé
ĐKXĐ : \(2x\ne3y\)
\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)
\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)
\(=\dfrac{\left(2x-3y\right)^3}{\left(2x-3y\right)^2}=2x-3y\)
Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3
a) \(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)
\(A=20x^3-10x^2+5x-20x^3+10x^2+4x\)
\(A=9x\)
Thay x = 15 vào, ta có:
\(A=9.15=135\)
b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(B=5x^2-20xy-4y^2+20xy\)
\(B=5x^2-4y\)
Thay \(x=-\frac{1}{5};y=-\frac{1}{2}\) vào, ta có:
\(B=5.\left(-\frac{1}{5}\right)^2-4.\left(-\frac{1}{2}\right)=\frac{11}{5}\)
c) \(C=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)-5y^2\left(x^2-xy\right)\)
\(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(C=9x^2y^2-xy^3-8x^3\)
Thay \(x=\frac{1}{2};y=2\) vào, ta có:
\(C=9.\left(\frac{1}{2}\right)^2.2^2-\frac{1}{2}.2^3-8.\left(\frac{1}{2}\right)^3=4\)
d) \(D=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)
\(D=6x^2-3x+10x-5+12x^2+8x-3x-2\)
\(D=18x^2+12x-7\)
Ta có: \(\left|2\right|=\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)
+) Với x = -2
\(D=18.\left(-2\right)^2+12.\left(-2\right)-7=41\)
+) Với x = 2
\(D=18.2^2+12.2-7=89\)
(4x - 1)(3 - x) - (2x + 3)(2x - 1) = 0
12x + \(4x^2\)- 3 + x - 4x + 2x - 6x + 3 = 0
\(4x^2\)+ 5x = 0
4x . 4x + 5x = 0
16x + 5x = 0
21x = 0
x = 0
2x(8x + 3) - (4x - 1)(4x + 1) = 13
16x + 6x - 16x - 4x + 4x +1 = 13
6x +1 = 13
6x = 12
x = 2