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6 tháng 6 2017

\(a,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)

\(=2x^2+2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3\left(x^2+y^2\right)\)\(b,\left(5x-1\right)+2\left(1-5x\right)\left(4x+5\right)+\left(5x+4\right)\)\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2=25\)

6 tháng 6 2017

c)\(Q=\left(x-y\right)^3+\left(x+y\right)^3+\left(x-y\right)^3-3xy\left(x+y\right)\)

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

\(=x^3+y^3\)

d)\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(2P=5^{32}-1\Rightarrow P=\dfrac{5^{32}-1}{2}\)

26 tháng 5 2017

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

Q=\(x^3+y^3\)

26 tháng 5 2017

P=\(\left(5x-1-5x-4\right)^2\)

P=25

10 tháng 7 2017

a, \(P=\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)

\(=5x-1+2\left(4+5x-20x-25x^2\right)+\left(25x^2+40x+16\right)\)

\(=5x-1+8-30x-50x^2+25x^2+40x+16\)

\(=\left(-50x^2+25x^2\right)+\left(5x-30x+40x\right)+\left(-1+8+16\right)\)

\(=-25x^2+15x+23\)

b, \(Q=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)

\(=x^3-3x^2y+3xy^2-y^3+y^3+3y^2x+3yx^2+x^3+y^3-3y^2x+3yx^2-x^3-3x^2y-3xy^2\)

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

\(=x^3+y^3\)

Chúc bạn học tốt!!!

3 tháng 7 2016

a)P=2(1-5x)(4+5x)+(5x+4)2

=5x-1-50x2-30x+8+25x2+40x+16

=(-50x2+25x2)+(5x-30x+40x)+8+16-1

=-25x2+15x+23

b)Q=(x-y)3+(y+x)3+(y-x)3-3xy(x+y)

=(x+y)3-3xy(x+y)

=x3+y3

a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)

\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)

b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)

\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)

=1/5-1=-4/5

\(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)

d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)

\(=20x^3-30x^2+15x+4\)

\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)

25 tháng 7 2017

Câu 1 :

\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)=\left(2x\right)^3+y^3=8x^3+y^3\)Câu 2:

\(A=3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)\(\Leftrightarrow3\left(6x^2-2x-6\right)-2\left(4x^2+13x-12\right)+36x-9x^2=0\)\(\Leftrightarrow18x^2-6x-18-8x^2-26x+24+36x-9x^2=0\)\(\Leftrightarrow x^2+4x+6=0\)

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

Ta có:

\(\left(x+2\right)^2\ge0\forall x\)

Vậy pt vô nghiệm

Vậy:ko......

Câu 3:

\(\left(5x-3\right)\left(7x+2\right)-35x\left(x-1\right)=42\)

\(\Leftrightarrow35x^2+10x-21x-6-35x^2+35x-42=0\)\(\Leftrightarrow14x=48\Leftrightarrow x=\dfrac{7}{24}\)

Câu 4:

\(\left(3x+5\right)\left(2x-1\right)+\left(5-6x\right)\left(x+2\right)=x\)

\(\Leftrightarrow6x^2-3x+10x-5+5x+10-6x^2-12x-x=0\)\(\Leftrightarrow-x=-5\Rightarrow x=5\)

câu 6,

25 tháng 7 2017

Câu 6: \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(\Rightarrow10x^2+9x-\left(10x^2-2x+15x-3\right)=8\)

\(\Rightarrow10x^2+9x-10x^2+2x-15x+3=8\)

\(\Rightarrow-4x+3=8\)

\(\Rightarrow-4x=5\Rightarrow x=\dfrac{-5}{4}\)

Câu 7: \(x\left(x+1\right)\left(x+6\right)-x^3=5x\)

\(\Rightarrow\left(x^2+x\right)\left(x+6\right)-x^3=5x\)

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

\(\Rightarrow7x^2=-x\)

\(\Rightarrow7x=-1\Rightarrow x=\dfrac{-1}{7}\).

9 tháng 7 2017

Thao Nguyen VT= Vế trái

VP= Vế phải

9 tháng 7 2017

2. CMR:

a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)

Ta có: VT=\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5=x^5-y^5=VP\)=> đpcm.

b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5+y^5\)

Ta có: VT=\(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5=VP\)

=> đpcm.

c. \(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)

\(\Leftrightarrow x^2+bx+ax+ab=x^2+ax+bx+ab\) (đúng)

=> đpcm.

17 tháng 8 2020

a) \(5x^2-2x\left(3x+\frac{3}{2}\right)=-x^2-3x=-x\left(x+3\right)=-3\left(3+3\right)=-18\)

b) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2=3\left(x^2-\frac{4}{5}y^2\right)\)

\(=3\left(4^2-\frac{4}{5}.5^2\right)=3.\left(-4\right)=-12\)

c) \(\left(x-2\right)^2-\left(x+7\right)\left(x-7\right)=x^2-4x+4-x^2+49=-4x+53=-4.3+53=41\)

d) \(x^2+12x+36=\left(x+6\right)^2=\left(64+6\right)^2=70^2=4900\)

e) \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=x^2-6x+9-x^2+16=-6x+25=-6\left(-1\right)+25\)

= 31

f) \(\left(3x+2y\right)^2-4y\left(3x+y\right)=9x^2+12xy+4y^2-12xy-4y^2=9x^2=9\left(-\frac{1}{3}\right)^2=1\)

17 tháng 8 2020

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

Thay x = 3 vào biểu thức trên ta cs : \(-3^2-3.3=-9-9=-18\)

b, \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2\)

Thay x = 4 ; y = 5 vào biểu thức trên ta có : \(3.4^2-\frac{12}{5}.5^2=-12\)