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

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

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

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

Gọi vận tốc riêng của ca là x ( x > 0 ) 

vận tốc ca nô xuôi dòng : x + 4 km/h 

vận tốc ca nô ngược dòng : x - 4 km/h 

Thời gian đi xuôi A đến B : \(\dfrac{30}{x+4}\)giờ

Thời gian đi từ B ngược về một điểm cách B 20 km nên ta có : \(\dfrac{20}{x-4}\)giờ

Tổng thời gian đi hết 2h30p = 5/2 h

Ta có pt \(\dfrac{30}{x+4}+\dfrac{20}{x-4}=\dfrac{5}{2}\Leftrightarrow x=20\)km/h 

   x2 + y2 + x2y2 + 1 + 4xy + 2(x + y)(1 + xy) = 25

<=> (x + y)2 + (xy + 1)2 + 2(x + y)(1 + xy) = 25

<=> [(x + y) + (xy +1)]2 = 25 

<=> [(x + 1)(y + 1)]2 = 25

<=> (x + 1)(y + 1) = 5    (x,y > 0)

Đáp số: (x = 4, y = 0) và (x = 0, y = 4)

@Kẻ mạo danh lần sau bạn có lấy bài giải từ nguồn khác thì ít nhất cũng phải ghi "Tham khảo" nhé, không thì tốt nhất bạn vẫn nên tự giải.

a, Cho O là giao điểm AC;BD 

Theo bđt tam giác 

OA + OB > AB ; OC + OD > DC

Cộng vế vs vế : OA + OB + OC + OD = AC + BD > AB + CD 

a: Xét ΔAEC vuông tại E và ΔAHB vuông tại H có

\(\widehat{EAC}\) chung

Do đó: ΔAEC~ΔAHB

b; Xét ΔHCB vuông tại H và ΔKAC vuông tại K có

\(\widehat{HCB}=\widehat{KAC}\)(AD//BC)

Do đó: ΔHCB~ΔKAC

=>\(\dfrac{HC}{AK}=\dfrac{BC}{CA}\)

=>\(BC\cdot AK=CH\cdot CA\)

c: Xét ΔBHA vuông tại H có \(sinBAH=\dfrac{BH}{BA}\)

=>\(\dfrac{2}{BA}=sin30=\dfrac{1}{2}\)

=>BA=4(cm)

ΔAHB~ΔAEC

=>\(\dfrac{S_{AHB}}{S_{AEC}}=\left(\dfrac{AB}{AC}\right)^2=\left(\dfrac{4}{3}\right)^2=\dfrac{16}{9}\)

\(10x^2+y^2+4z^2+6x-4y-4xz=-5\\ =>10x^2+y^2+4z^2+6x-4y-4xz+5=0\\ =>\left(9x^2+6x+1\right)+\left(x^2-4xz+4z^2\right)+\left(y^2-4y+4\right)=0\\ =>\left(3x+1\right)^2+\left(x-2z\right)^2+\left(y-2\right)^2=0\)

Mà: \(\left\{{}\begin{matrix}\left(3x+1\right)^2\ge0\forall x\\\left(x-2z\right)^2\ge0\forall x,z\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.=>\left(3x+1\right)^2+\left(x-2z\right)^2+\left(y-2\right)^2\ge0\forall x,y,z\) 

\(=>\left\{{}\begin{matrix}3x+1=0\\x-2z=0\\y-2=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\z=-\dfrac{1}{6}\\y=2\end{matrix}\right.\)

\(10x^2+y^2+4z^2+6x-4y-4xz=-5\\ \Leftrightarrow\left(x^2-4xz+4z^2\right)+\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)=0\\ \Leftrightarrow\left(x-2z\right)^2+\left(3x+1\right)^2+\left(y-2\right)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2z\right)^2\ge0\forall x,z\\\left(3x+1\right)^2\ge0\forall x\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2z\right)^2+\left(3x+1\right)^2+\left(y-2\right)^2\ge0\forall x,y,z\)

Mà: \(\left(x-2z\right)^2+\left(3x+1\right)^2+\left(y-2\right)^2=0\)

Do đó: \(\left\{{}\begin{matrix}x-2z=0\\3x+1=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=2\\z=-\dfrac{1}{6}\end{matrix}\right.\)

#$\mathtt{Toru}$

Bài 8:

1: \(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=2x\cdot2y=4xy\)

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

\(=4x^2+12x+9-6x^2-3x\)

\(=-2x^2+9x+9\)

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

\(=4^2-\left(2x\right)^2-8x^2-12x\)

\(=16-4x^2-8x^2-12x=-12x^2-12x+16\)

4: \(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2-2x^2\)

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

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

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

\(=9x^2-6x+12x-8-9x^2-6x-1\)

=-9

6: \(4x\left(x-3\right)-\left(2x-1\right)\left(2x+1\right)\)

\(=4x^2-12x-\left(4x^2-1\right)\)

\(=4x^2-12x-4x^2+1=-12x+1\)

7: \(\dfrac{3}{2}x^2-\left(x-1\right)\left(x+1\right)+3x\)

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

\(=\dfrac{3}{2}x^2+3x-x^2+1=\dfrac{1}{2}x^2+3x+1\)

8: \(2\left(5-x\right)\left(5+x\right)-\left(2x+3\right)^2-x\left(3x+2\right)\)

\(=2\left(25-x^2\right)-4x^2-12x-9-3x^2-2x\)

\(=2\left(25-x^2\right)-7x^2-14x-9\)

\(=50-2x^2-7x^2-14x-9=-9x^2-14x+41\)

Giúp tớ nhanh vs ạ

Bài 8:

\(1)\left(x+y\right)^2-\left(x-y\right)^2\\ =\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\\ =x^2+2xy+y^2-x^2+2xy-y^2\\ =4xy\\ 2)\left(2x+3\right)^2-3x\left(2x+1\right)\\ =\left(4x^2+12x+9\right)-\left(6x^2+3x\right)\\ =4x^2+12x+9-6x^2-3x\\ =-2x^2+9x+9\\ 3)\left(4-2x\right)\left(4+2x\right)-4x\left(2x+3\right)\\ =\left[4^2-\left(2x\right)^2\right]-\left(8x^2+12x\right)\\ =16-4x^2-8x^2-12x\\ =16-12x^2-12x\\ 4)2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2-2x^2\\ =2\left(x^2-y^2\right)+\left(x^2+2xy+y^2\right)-2x^2\\ =2x^2-2y^2+x^2+2xy+y^2-2x^2\\ =x^2+2xy-y^2\)

Bài 8:

1: \(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=2x\cdot2y=4xy\)

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

\(=4x^2+12x+9-6x^2-3x\)

\(=-2x^2+9x+9\)

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

\(=4^2-\left(2x\right)^2-8x^2-12x\)

\(=16-4x^2-8x^2-12x=-12x^2-12x+16\)

4: \(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2-2x^2\)

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

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

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

\(=9x^2-6x+12x-8-9x^2-6x-1\)

=-9

6: \(4x\left(x-3\right)-\left(2x-1\right)\left(2x+1\right)\)

\(=4x^2-12x-\left(4x^2-1\right)\)

\(=4x^2-12x-4x^2+1=-12x+1\)

7: \(\dfrac{3}{2}x^2-\left(x-1\right)\left(x+1\right)+3x\)

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

\(=\dfrac{3}{2}x^2+3x-x^2+1=\dfrac{1}{2}x^2+3x+1\)

8: \(2\left(5-x\right)\left(5+x\right)-\left(2x+3\right)^2-x\left(3x+2\right)\)

\(=2\left(25-x^2\right)-4x^2-12x-9-3x^2-2x\)

\(=2\left(25-x^2\right)-7x^2-14x-9\)

\(=50-2x^2-7x^2-14x-9=-9x^2-14x+41\)