-Làm tính nhân:a. \(x^2\left(x-2x^3\right)\)b. \(\left(x-2\right)\left(x-x^2 4\right)\)c. \(\left(x^2-1\right)\left(x^2 2x\right)\)d. \(\left(2x-1\right)\left(3x 2\right)\left(3-x\right)\)e. \(\left(x...

DV

Đàm Vũ Long

16 tháng 6 2018 - olm

-Làm tính nhân:a. \(x^2\left(x-2x^3\right)\)b. \(\left(x-2\right)\left(x-x^2+4\right)\)c. \(\left(x^2-1\right)\left(x^2+2x\right)\)d. \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)e. \(\left(x+3\right)\left(x^2+3x-5\right)\)f. \(\left(xy-2\right)\left(x^3-2x-6\right)\)g. \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)GIÚP MK NHÉ SẮP PHẢI NỘP BÀI...

Đọc tiếp

#Toán lớp 8

2

D

Dương

16 tháng 6 2018

\(a,x^2\left(x-2x^3\right)\)

\(=x^3-2x^5\)

\(b,\left(x-2\right)\left(x-x^2+4\right)\)

\(=x^2-x^3+4x-2x+2x^2-8\)

\(=3x^2-x^3+2x-8\)

\(c,\left(x^2-1\right)\left(x^2+2x\right)\)

\(=x^4+2x^3-x^2-2x\)

\(d,\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)

\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)

\(=\left(6x^2+x-2\right)\left(3-x\right)\)

\(=18x^2+3x-6-6x^3-x^2+2x\)

\(=17x^2+5x-6-6x^3-x^2\)

\(e,\left(x+3\right)\left(x^2+3x-5\right)\)

\(=x^3+3x^2-5x+3x^2+9x-15\)

\(=x^3+6x^2+4x-15\)

\(f,\left(xy-2\right)\left(x^3-2x-6\right)\)

\(=x^4y-2x^2y-6xy-2x^3+4x-12\)

\(g,\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)

\(=20x^5-4x^4+8x^3-12x^2-5x^4+x^3-2x^2+3x+10x^3-2x^2+4x-6\)

\(=20x^5-9x^4+19x^3-16x^2+7x-6\)

Đúng(0)

VL

VRCT_Ran Love Shinichi

16 tháng 6 2018

a. x²(x−2x³)= x³-2x⁵

b. (x−2)(x−x²+4)= x²-x³+4x-2x+2x²-8= -x³+3x²+2x-8

c. (x²−1)(x²+2x)= x⁴+2x³-x²-2x

d. (2x−1)(3x+2)(3−x) = (6x²+x-2)(3-x)=18x²-6x³+3x-x²-6+2x =-6x³+17x²+5x-6

e. (x+3)(x²+3x−5)= x³+3x²-5x+3x²+9x-15= x³+6x²+4x-15

f. (xy−2)(x³−2x−6)= x⁴y-2x²y-6xy-2x³+4x+12

g. (5x³−x²+2x−3)(4x²−x+2)= 20x⁵-9x⁴+19x³-12x²+7x-6

Đúng(0)

NH

Nhi huỳnh

20 tháng 8 2021

d) \(^{ }4x\left(2x+3\right)-8x\left(x+4\right)\)

e) \(^{ }2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)

f) \(^{ }x\left(x+2\right)^2-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

20 tháng 8 2021

d: Ta có: \(4x\left(2x+3\right)-8x\left(x+4\right)\)

\(=8x^2+12x-8x^2-32x\)

=-20x

e: Ta có: \(2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)

\(=10x^2+4x+6x^2-2x-9x+3\)

\(=16x^2-7x+3\)

f: Ta có: \(x\left(x+2\right)^2-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

\(=x^3+4x^2+4x-x^3-3x^2-3x-1+3x^2-3\)

\(=4x^2+x-4\)

Đúng(1)

NN

nguyen ngoc son

16 tháng 2 2021

giải phương trình

a.\(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)

b.\(x\left(2x-9\right)=3x\left(x-5\right)\)

c.\(3x-15=2x\left(x-5\right)\)

d.\(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)

e.\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

16 tháng 2 2021

a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)

b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)

\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)

\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)

\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)

\(\Leftrightarrow x\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: S={0;6}

c) Ta có: \(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)

d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)

\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)

\(\Leftrightarrow30-6x=6x-8\)

\(\Leftrightarrow30-6x-6x+8=0\)

\(\Leftrightarrow-12x+38=0\)

\(\Leftrightarrow-12x=-38\)

\(\Leftrightarrow x=\dfrac{19}{6}\)

Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)

e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow6x+4-3x-1=12x+10\)

\(\Leftrightarrow3x+3-12x-10=0\)

\(\Leftrightarrow-9x-7=0\)

\(\Leftrightarrow-9x=7\)

\(\Leftrightarrow x=-\dfrac{7}{9}\)

Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)

Đúng(2)

BT

Bùi Thanh Tâm

25 tháng 1 2021

Giair phương trình sau:a,\(2x^3+5x^2-3x=0\) b,\(2x^3+6x^2=x^2+3x\)c,\(x^2+\left(x+2\right)\left(11x-7\right)=4\) d,\(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)e, \(x^3+1=x\left(x+1\right)\) f,\(x^3+x^2+x+1=0\)g,\(x^3-3x^2+3x-1=0\) h,\(x^3-7x+6=0\)i,\(x^6-x^2=0\) j,\(x^3-12=13x\)k,\(-x^5+4x^4=-12x^3\) ...

Đọc tiếp

#Toán lớp 8

2

NL

Nguyễn Lê Phước Thịnh

25 tháng 1 2021

a) Ta có: \(2x^3+5x^2-3x=0\)

\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)

\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)

\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)

\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)

b) Ta có: \(2x^3+6x^2=x^2+3x\)

\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)

\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)

c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)

\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)

\(\Leftrightarrow12x^2+15x-18=0\)

\(\Leftrightarrow12x^2+24x-9x-18=0\)

\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)

Đúng(2)

KD

Kieu Diem

25 tháng 1 2021

Trong đó có nhiều phương trình kiến thức cơ bản mà nhỉ? Ít nâng cao, bạn lọc ra câu nào k làm đc thôi chứ!

Đúng(1)

HH

Hồ Hữu Duyy

26 tháng 12 2021

Thực hiện phép tính:

a) \(2x.\left(2x^2+3x-1\right)\)

b) \(\left(x+5\right).\left(2x-3\right)\)

c) \(\left(x+1\right)^2-x\left(2+3x\right)\)

d) \(\left(2x^3+x^2-8x+3\right):\left(2x-3\right)\)

#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

26 tháng 12 2021

b: \(=2x^2-3x+10x-15=2x^2+7x-15\)

Đúng(0)

NK

Ng KimAnhh

25 tháng 2 2023

Tính:

\(a)\left(-2x^2\right)\cdot\left(3x-4x^3+7-x^2\right)\)

\(b)\left(x+3\right)\cdot\left(2x^2-3x-5\right)\)

\(c)\left(-6x^5+7x^4-6x^3\right):3x^3\)

\(d)\left(9x^2-4\right):\left(3x+2\right)\)

\(e)\left(2x^4-13x^3+15x^2+11x-3\right):\left(x^2-4x-3\right)\)

#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

25 tháng 2 2023

a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)

\(=8x^5+2x^4-6x^3-14x^2\)

b: \(=2x^3-3x^2-5x+6x^2-9x-15\)

\(=2x^3+3x^2-14x-15\)

c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)

d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)

e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)

=2x^2-5x+1

Đúng(2)

NM

Ngọc Minh

8 tháng 9 2023

tìm GTLN

a)\(A=x^2+5y^2+2xy-4x-8y+2015\)

b)\(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)

c)\(C=\left(x-1\right)\left(2x-1\right)\left(2x^2-3x-1\right)+2017\)

d)\(D=\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+10\)

#Toán lớp 8

3

NL

Nguyễn Lê Phước Thịnh

8 tháng 9 2023

loading...

Đúng(0)

NT

Nguyễn thành Đạt

CTVHS

8 tháng 9 2023

Bạn xem lại đề nhé.

a) \(A=x^2+5y^2+2xy-4x-8y+2015\)

\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2-y\right)^2+4y^2+2011\)

Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)

\(\Rightarrow A_{min}=2011\)

Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

Đúng(0)

MT

Mai Thị Thúy

22 tháng 7 2021

giải pt :

a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)

b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)

c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)

d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)

#Toán lớp 11

0