`#3107.101107`

`c,`

`(x^3)/2 + 4`

`= (x^3)/2 + 8/2`

`= (x^3 + 8)/2`

`= 1/2*(x^3 + 8)`

`= 1/2*(x + 2)(x^2 - 2x + 4)`

`d,`

`27y^3 + 27y^2 + 9y + 1`

`= (3y)^3 + 3 * (3y)^2 * 1 + 3 * 3y * 1^2 + 1^3`

`= (3y + 1)^3`

____

HĐT:

`A^3 + B^3 = (A + B)(A^2 - AB + B^2)`

`(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3.`

Gọi bìa sách có giá thấp là: `x` (đồng)

ĐK: `x>0` 

Bìa sách có giá cao là: `x+15000` (đồng) 

Hoàng phải trả số tiền khi mua bìa sách giá thấp là: `(100%-20%)x=0,8x` (đồng)

Hoàng phải trả số tiền khi mua bìa sách giá cao là: `(100%-20%)(x+15000)=0,8(x+15000)` (đồng) 

Mà tổng giá tiền Hoàng phải trả cho cô thu ngân là 84000 đồng nên ta có pt: 

\(0,8x+0,8\left(x+15000\right)=84000\\\Leftrightarrow0,8x+0,8x+12000=84000\\ \Leftrightarrow1,6x+12000=84000\\ \Leftrightarrow1,6x=84000-12000\\ \Leftrightarrow1,6x=72000\\ \Leftrightarrow x=72000:1,6\\ \Leftrightarrow x=45000\left(tm\right)\)

Bìa sách giá thấp có giá là 45000 (đồng)

Bìa sách giá cao có giá là: 45000 + 15000 = 60000 (đồng) 

a: Đặt \(A=\left(2x+y\right)^2-2\left(2x+y\right)\left(2x-y\right)+\left(2x-y\right)^2\)

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

Khi y=3 thì \(A=4\cdot3^2=4\cdot9=36\)

b: Đặt \(B=\left(2x-5\right)\left(2x+5\right)-\left(2x+1\right)^2\)

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

\(=4x^2-25-4x^2-4x-1=-4x-26\)

Khi x=0 thì \(B=-4\cdot0-26=-26\)

a: Đặt \(A=\left(2x+y\right)^2-2\left(2x+y\right)\left(2x-y\right)+\left(2x-y\right)^2\)

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

Khi y=3 thì \(A=4\cdot3^2=4\cdot9=36\)

b: Đặt \(B=\left(2x-5\right)\left(2x+5\right)-\left(2x+1\right)^2\)

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

\(=4x^2-25-4x^2-4x-1=-4x-26\)

Khi x=0 thì \(B=-4\cdot0-26=-26\)

3 lần số bé là:

             1006-124=882

Số bé là:

             882:3=294

Số lớn là:

         294*2+124=712

gọi 2 số đó là a,b (a,b là số tự nhiên, a>b)

Theo bài ra thì a+b=1006 và a chia b được 2 dư 124 

=>(a-124)/b=2 => a-124=2b =>a=124+2b

mà a+b=1006

suy ra 124+2b+b=1006 => b=294; a=712

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

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

#$\mathtt{Toru}$

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

=>\(2x^2-3x-6+2x^2+1=0\)

=>\(4x^2-3x-5=0\)

\(\text{Δ}=\left(-3\right)^2-4\cdot4\cdot\left(-5\right)=9+80=89>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}x=\dfrac{3-\sqrt{89}}{2\cdot4}=\dfrac{3-\sqrt{89}}{8}\\x=\dfrac{3+\sqrt{89}}{2\cdot4}=\dfrac{3+\sqrt{89}}{8}\end{matrix}\right.\)

Bài 4:

\(a)2,6^2+4\cdot1,3\cdot7,4+7,4^2\\ =2,6^2+2\cdot\left(2\cdot1,3\right)+7,4^2\\ =2,6^2+2\cdot2,6\cdot7,4+7,4^2\\ =\left(2,6+7,4\right)^2\\ =10^2\\ =100\\ b)2024^2-2023^2\\ =\left(2024-2023\right)\left(2024+2023\right)\\ =1\cdot4047\\ =4047\)

Bài 5:

\(a)4x^2+24x+36\\ =\left(2x\right)^2+2\cdot2x\cdot6+6^2\\ =\left(2x+6\right)^2\\ b)9x^4y^2+18x^2y+9\\ =\left(3x^2y\right)^2+2\cdot3x^2y\cdot3+3^2\\ =\left(3x^2y+3\right)^2\)

giải giúp mình bài 5 với các bạn

Bài 13:

\(1)A=x^2-x+1\\ =\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\\ 2)B=x^2+x+1\\ =\left(x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\\ 3)C=x^2+2x+2\\ =\left(x^2+2x+1\right)+1\\ =\left(x+1\right)^2+1\ge1>0\forall x\)

\(4)A=x^2-5x+10\\ =\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}\right)+\dfrac{15}{4}\\ =\left(x-\dfrac{5}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}>0\forall x\\ 5)B=x^2-8x+20\\ =\left(x^2-8x+16\right)+4\\ =\left(x-4\right)^2+4\ge4>0\forall x\\ 6)C=x^2-8x+17\\ =\left(x^2-8x+16\right)+1\\ =\left(x-4\right)^2+1\ge1>0\forall x\) 

\(7)A=x^2-6x+10\\ =\left(x^2-6x+9\right)+1\\ =\left(x-3\right)^2+1\ge1>0\forall x\\ 8)B=9x^2-6x+2\\ =\left(9x^2-6x+1\right)+1\\ =\left(3x-1\right)^2+1\ge1>0\forall x\\ 9)C=2x^2+8x+15\\ =\left(2x^2+8x+8\right)+7\\ =2\left(x^2+4x+4\right)+7\\ =2\left(x+2\right)^2+7\ge7>0\forall x\)

\(a)\left(x+5\right)^2=x^2+2\cdot x\cdot5+5^2=x^2+10x+25\\ b)\left(2-y\right)^2=2^2-2\cdot2\cdot y+y^2=4-4y+y^2\\ c)\left(5x-1\right)^2=\left(5x\right)^2-2\cdot5x\cdot1+1^2=25x^2-10x+1\\ d)\left(1+5x^3\right)^2=1^2+2\cdot1\cdot5x^3+\left(5x^3\right)^2=1+10x^3+25x^6\\ e)\left(7-a^2\right)\left(7+a^2\right)=7^2-\left(a^2\right)^2=49-a^4\\ \left(x-y\right)^2-\left(x+y\right)^2=\left(x-y+x+y\right)\left(x-y-x-y\right)=2x\cdot-2y=-4xy\\ g)\left(2x^3-\dfrac{1}{2}y\right)^2=\left(2x^3\right)^2-2\cdot2x^3\cdot\dfrac{1}{2}y+\left(\dfrac{1}{2}y^2\right)=4x^6-2x^3y+\dfrac{1}{4}y^2\\ h)\left(x^2+4y\right)^2=\left(x^2\right)^2+2\cdot x^2\cdot4y+\left(4y\right)^2=x^4+8x^2y+16y^2\\ i)\left(a+b+c\right)^2=\left[a+\left(b+c\right)\right]^2=a^2+2a\left(b+c\right)+\left(b+c\right)^2\\ =a^2+2ab+2ac+b^2+2bc+c^2=a^2+b^2+c^2+2ab+2bc+2ac\\ k)\left(a-b-c\right)^2=\left[a-\left(b+c\right)\right]^2=a^2-2a\left(b+c\right)+\left(b+c\right)^2\\ =a^2-2ab-2ac+b^2+2bc+c^2=a^2+b^2+c^2-2ab-2ac+2bc\)