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25 tháng 2 2019

Đề lỗi rồi kìa ba: \(+^2+\) là sao?

28 tháng 2 2019

Khó ghê,giải giúp anh với :v

a) Ta có: \(-2xy^2\cdot\left(x^3y-2x^2y^2+5xy^3\right)\)

\(=-2x^4y^3+4x^3y^4-10x^2y^5\)

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

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

c) Ta có: \(3x^2\left(2x^3-x+5\right)\)

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

d) Ta có: \(\left(-10x^3+\frac{2}{5}y-\frac{1}{3}z\right)\cdot\left(-\frac{1}{2}xy\right)\)

\(=5x^4y-\frac{1}{5}xy^2+\frac{1}{6}xyz\)

e) Ta có: \(\left(3x^2y-6xy+9x\right)\cdot\left(-\frac{4}{3}xy\right)\)

\(=-4x^3y^2+8x^2y^2-12x^2y\)

f) Ta có: \(\left(4xy+3y-5x\right)\cdot x^2y\)

\(=4x^3y^2+3x^2y^2-5x^3y\)

5 tháng 8 2020

a)

\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)

\(=x^3-3x^2+9x+3x^2-9x+27-54-x^3\)

\(=-27\)

or

\(A=x^3+27-54-x^3=-27\)

b)

\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

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

c)

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

\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)

d)

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

\(=x^3-8-\left(x-1\right)^3+3\left(x-1\right)\left(x+1\right)\)

\(=6x^2-3x-10\)

a) Ta có: \(5x^2-3x\left(x+2\right)\)

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

\(=2x^2-6x\)

b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)

\(=3x^2-15x-5x^2-35x\)

\(=-2x^2-50x\)

c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)

\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)

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

d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)

\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)

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

\(=-4x^2y+5x^2-2x\)

e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)

\(=4x^4-16x^3+4x^4-2x^3+14x^2\)

\(=8x^4-18x^3+14x^2\)

f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)

\(=25x-12x+4+35x-14x^3\)

\(=-14x^3+48x+4\)

3 tháng 3 2022

\(a.ĐK:x\ne3;1\)

\(\Rightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)+2\left(3x-10\right)}{2\left(x-1\right)\left(x-3\right)}=\dfrac{7\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow x-1+6x-20=7\left(x^2-4x+3\right)\)

\(\Leftrightarrow7x-21=7x^2-28x+21\)

\(\Leftrightarrow7x^2-35x+42=0\)

\(\Leftrightarrow7\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow x^2-5x+6=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)

b.\(ĐK:x\ne2;4\)

\(\Rightarrow\dfrac{x-1}{x-2}-\dfrac{x+3}{4-x}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(4-x\right)}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)

\(\Leftrightarrow\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)

\(\Leftrightarrow4x-x^2-4+x-x^2+2x-3x+6-2=0\)

\(\Leftrightarrow-2x^2+4x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)

 

 

a: \(\Leftrightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)

\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow7\left(x^2-4x+3\right)=x-1+6x-20=7x-21\)

\(\Leftrightarrow\left(x-3\right)\left(7x-7\right)-7\left(x-3\right)=0\)

=>(x-3)(7x-14)=0

=>x=3(loại) hoặc x=2(nhận)

b: \(\Leftrightarrow\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)=-2\)

\(\Leftrightarrow x^2-5x+4+x^2+x-6=-2\)

\(\Leftrightarrow2x^2-4x=0\)

=>2x(x-2)=0

=>x=0(nhận) hoặc x=2(loại)

27 tháng 11 2020

Bài 1 : 

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

Mà \(B=-\left(y^2-x\right)^2\)

Nên ta có : đpcm 

27 tháng 11 2020

Bài 2 

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

TH1 : x = -1

TH2 : x = 2

TH3 : x = 1/2 

Bài 4 : 

a, \(\left(2x+3\right)\left(5-x\right)=0\Leftrightarrow x=-\frac{3}{2};5\)

b, \(\left(x-\frac{1}{2}\right)\left(3x+1\right)\left(2-x\right)=0\Leftrightarrow x=\frac{1}{2};-\frac{1}{3};2\)

c, \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;-2\)

d, \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow x=0;1\)

13 tháng 3 2019

\(\left(-2xy^2\right)\left(3x^3yz\right)\)

\(=-2xy^2.3x^3yz\)

\(=\left(-2.3\right)\left(xx^3\right)\left(y^2y\right)z\)

\(=-6x^4y^3z\)

13 tháng 3 2019

a)\(=-6x^4y^3z\)

b)\(=\left(-2x^2y\right)^2.\left(3xy^3\right)^2.\frac{1}{24}x\)

\(=4x^4y^2.9x^2y^6.\frac{1}{24}x\)

\(=\frac{3}{2}x^5y^8\)

24 tháng 4 2023

\(Q\left(x\right)=-3x^4+4x^3+2x^2+\dfrac{2}{3}-3x-2x^4-4x^3+8x^4+1+3x\)
\(=\left(-3x^4-2x^4+8x^4\right)+\left(4x^3-4x^3\right)+2x^2-\left(3x-3x\right)+\left(1+\dfrac{2}{3}\right)\)
\(=3x^4+2x^2+\dfrac{5}{3}\)
\(3x^4+2x^2+\dfrac{5}{3}=0\)
\(\Rightarrow3x^4+2x^2=-\dfrac{5}{3}\)(Vô lí vì \(3x^4\) và \(2x^2\) luôn lớn hơn hoặc bằng 0)
Vậy Q(x) không có nghiệm

Q(x)=3x^4+2x^2+5/3>=5/3>0 với mọi x

=>Q(x) vô nghiệm