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a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)
Thay x = 15 vào bt A ta có
A = 9 . 15 = 135
b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)
Thay x = -1/5 ; y = - 1/2 vào bt B ta có
\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(=9x^2y^2-xy^3-8x^3\)
Thay x = 1/2 ; y = 2 vào bt C ta có
\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)
d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)
\(=12x^2+12x-3\)
\(\left|x\right|=2\Rightarrow x=\pm2\)
Thay x = 2 vào bt D có
\(D=12.4+12.2-3=69\)
Thay x = - 2 vào bt D ta có
\(D=12.4-12.2-3=21\)
Lời giải:
a.
$A=20x^3-10x^2+5x-(20x^3-10x^2-4x)$
$=9x=9.15=135$
b.
$B=(5x^2-20xy)-(4y^2-20xy)=5x^2-4y^2$
$=5(\frac{-1}{5})^2-4(\frac{-1}{2})^2=\frac{-4}{5}$
c.
$C=(6x^2y^2-6xy^3)-(8x^3-8x^2y^2)-(5x^2y^2-5xy^3)$
$=-8x^3+9x^2y^2-xy^3$
$=(-2x)^3+(3xy)^2-xy^3$
$=(-2.\frac{1}{2})^3+(3.\frac{1}{2}.2)^2-\frac{1}{2}.2^3$
$=(-1)^3+3^2-4=4$
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
\(D=18.2^2+12.2-7=89\)
1. a) \(( 5x-1)^2 - (5x-4) ( 5x+4) = 7\)
\(\Leftrightarrow\)\(25x^2-10x+1-(25x^2-16)=7\)
\(\Leftrightarrow\)\(25x^2-10x+1-25x^2+16-7=0\)
\(\Leftrightarrow\)\(10x=10\)
\(\Rightarrow x=1\)
b) \(( 4x-1)^2 - (2x+3)^2 + 5(x+2)^2 + 3(x-2) ( x+2) = 500\)
\(\Leftrightarrow\)\(16x^2-8x+1-4x^2-12x-9+5x+10+3x^2-12=500\)
\(\Leftrightarrow\)\(15x^2-15x=510\)
\(\Leftrightarrow\)\(15(x^2-x)=510\)
\(\Leftrightarrow\)\(x^2-x=34\)
\(\Rightarrow x=-5,352349955\)
c) \((x-2)^3 - (x-2) ( x^2+2x+4 ) + 6(x-2)(x+2) = 60\)
\(\Leftrightarrow x^3-6x^2+12x-8-\left(x^3-2^3\right)+6\left(x^2-4\right)=60\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+8+6x^2-24=60\)
\(\Leftrightarrow12x-24=60\)
\(\Leftrightarrow12x=84\)
\(\Rightarrow x=7\)
\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
Bài 1;
a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)
b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)
c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)
câu 1/ 5x(\(4x^2\)-2x+1) - 2x(\(10x^2\)-5x-2)
= 5x.\(4x^2\)-5x.2x+ 5x.1 - ( 2x.\(10x^2\)-2x.5x-2x.2)
= 9\(x^3\)-10\(x^2\)+5x - 20\(x^3\)+10\(x^2\)+4x
= (9\(x^3\)-\(20x^3\)) + (-10\(x^2\)+10\(x^2\)) + (5x+4x)
= \(-11x^3\) + 9x
à cj ơi, e 2k6, đọc phần lí thuyết r lm, nên có lỗi sai j mong cj thông cảm
a) Ta có: \(x\left(10x+9\right)-\left(5x-1\right)\left(2x+3\right)\)
\(=10x^2+9x-10x^2-15x+2x+3\)
\(=-4x+3\)
b) Ta có: \(\left(\dfrac{1}{3}-y^4\right)^2\)
\(=\left(\dfrac{1}{3}\right)^2-2\cdot\dfrac{1}{3}\cdot y^4+\left(y^4\right)^2\)
\(=y^8-\dfrac{2}{3}y^4+\dfrac{1}{9}\)
c) Ta có: \(\left(5x+y\right)^2\)
\(=\left(5x\right)^2+2\cdot5x\cdot y+y^2\)
\(=25x^2+10xy+y^2\)
a) (10x+9)x-(5x-1)(2x+3)
= 10x2+9x-10x2-15x+2x+3
= -4x+3