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

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

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

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

\(=5^2=25\)

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

\(=\left[\left(3x-5\right)-\left(3x+5\right)\right]^2\)

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

\(=\left(-10\right)^2\)

\(=100\)

a)\(9x^2+30x+25+9x^2-30x+25-\left(9x^2-2^2\right)\)

=\(9x^2+54\)=\(9\left(x^2+6\right)\)

b)\(2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)

=\(8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)

=\(x^3-16x^2+25x\)

c)\(\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2\)

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

Vì bài dài nên mình sẽ tách ra nhé.

1a. Ta có:

$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$

$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$

$=-3(-z)(-x)(-y)=3xyz$

$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$

------------------------

$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$

$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$

$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$

$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$

$=-z^5+5xyz^3-5x^2y^2z$

$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$

$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$

Từ $(1);(2)$ ta có đpcm.

1b.

$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$

$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$

$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$

Do đó:

$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$

$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$

$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$

$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$

$=7xyz(x^2y^2-2xyz^2+z^4)$

$=7xyz(xy-z^2)$

$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$

$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$

$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)

1c. Sử dụng kq phần a,b:

\(10(x^7+y^7+z^7)=70xyz(xy+yz+xz)^2\)

\(=-35xyz(xy+yz+xz).-2(xy+yz+xz)=-35xyz(x+y+z)(x^2+y^2+z^2)\)

\(=\frac{7}{6}.-30xyz(xy+yz+xz)(x^2+y^2+z^2)=\frac{7}{6}.6(x^5+y^5+z^5).(x^2+y^2+z^2)\)

\(=7(x^5+y^5+z^5)(x^2+y^2+z^5)\)

(đpcm)

1d. Áp dụng kq phần a
$6(x^5+y^5+z^5)=-30xyz(xy+y+xz)=15xyz.-2(xy+yz+xz)=15xyz(x^2+y^2+z^2)$

$\Rightarrow 2(x^5+y^5+z^5)=5xyz(x^2+y^2+z^2)$ (đpcm)

a)A=x³+x²y+y²x+y³+2x²y+2xy²

=x³+3x²y+3xy²+y³

A=(x+y)³

b)=3x²+2x+(x²+2x+1)-(4x²-25)=12

3x²+2x+x²+2x+1-4x²+25=12

4x+26=12

= >4x=6/13

= >x=6,5

\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)

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

\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\\ =3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\\=-10xy+2y^2 \)

b) \(3\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)

\(=3\left(4x^2+20x+25\right)-3\left(1+4x\right)\left(1-4x\right)\)

\(=12x^2+60x+75-3\left(1-16x^2\right)\)

\(=12x^2+60x+75-3+48x^2\)

\(=60x^2+60x+72\)

câu b là \(^{2\left(2x+5\right)^2-\left(4x+1\right)\cdot\left(1-4x\right)}\) xin lỗi mk chép đề sai ạ

1, \(A=5x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)

\(A=5x^3-15x+7x^2-5x^3-7x^2\)

\(A=\left(5x^3-5x^3\right)+\left(7x^2-7x^2\right)-15x\)

\(A=-15x\)

Thay \(x=-5\) vào A ta được:

\(-15\cdot-5=75\)

Vậy: ....

2. \(B=x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)

\(B=x^3-3x+7x^2-5x^3-7x^2\)

\(B=\left(x^3-5x^3\right)+\left(7x^2-7x^2\right)-3x\)

\(B=-4x^3-3x\)

Thay \(x=10,y=-1\) vào B ta được:

\(-4\cdot10^3-3\cdot10=-4\cdot1000-3\cdot10=-4000-30=-4030\)

Vậy: ....

Bài 1:

a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)

\(=10-10x=10(1-x)\)

b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)

\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)

\(=-7x^2+7x=7x(1-x)\)

c)

\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)

\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)

\(=\left\{3-x-5[9x-2]\right\}(-2x)\)

\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)

Bài 2:

a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)

\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)

\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)

b)

\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)

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

\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)

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

Bài 2:

c) \(2x(5-3x)+2x(3x-5)-3(x-7)=3\)

\(\Leftrightarrow 2x(5-3x)-2x(5-3x)-3(x-7)=3\)

\(\Leftrightarrow -3(x-7)=3\)

\(\Leftrightarrow x-7=-1\Rightarrow x=6\)

d)

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

\(\Leftrightarrow 3x^2+3x-(2x^2+4x)+x+1=0\)

\(\Leftrightarrow x^2+1=0\)

Vô lý vì \(x^2+1\geq 0+1=1>0\) với mọi $x$

Vậy không tồn tại $x$ thỏa mãn.

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy+y^2\)

\(\Leftrightarrow2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2\)

\(\Leftrightarrow4x^2\)