Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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:
\(a)4x^2+24x+36\\ =\left(2x\right)^2+2\cdot2x\cdot6+6^2\\ =\left(2x+6\right)^2\\ b)9x^4y^2+18x^2y+9\\ =\left(3x^2y\right)^2+2\cdot3x^2y\cdot3+3^2\\ =\left(3x^2y+3\right)^2\)
Bài 13:
\(1)A=x^2-x+1\\ =\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\\ 2)B=x^2+x+1\\ =\left(x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\\ 3)C=x^2+2x+2\\ =\left(x^2+2x+1\right)+1\\ =\left(x+1\right)^2+1\ge1>0\forall x\)
\(4)A=x^2-5x+10\\ =\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}\right)+\dfrac{15}{4}\\ =\left(x-\dfrac{5}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}>0\forall x\\ 5)B=x^2-8x+20\\ =\left(x^2-8x+16\right)+4\\ =\left(x-4\right)^2+4\ge4>0\forall x\\ 6)C=x^2-8x+17\\ =\left(x^2-8x+16\right)+1\\ =\left(x-4\right)^2+1\ge1>0\forall x\)
\(7)A=x^2-6x+10\\ =\left(x^2-6x+9\right)+1\\ =\left(x-3\right)^2+1\ge1>0\forall x\\ 8)B=9x^2-6x+2\\ =\left(9x^2-6x+1\right)+1\\ =\left(3x-1\right)^2+1\ge1>0\forall x\\ 9)C=2x^2+8x+15\\ =\left(2x^2+8x+8\right)+7\\ =2\left(x^2+4x+4\right)+7\\ =2\left(x+2\right)^2+7\ge7>0\forall x\)
\(a)\left(x+5\right)^2=x^2+2\cdot x\cdot5+5^2=x^2+10x+25\\ b)\left(2-y\right)^2=2^2-2\cdot2\cdot y+y^2=4-4y+y^2\\ c)\left(5x-1\right)^2=\left(5x\right)^2-2\cdot5x\cdot1+1^2=25x^2-10x+1\\ d)\left(1+5x^3\right)^2=1^2+2\cdot1\cdot5x^3+\left(5x^3\right)^2=1+10x^3+25x^6\\ e)\left(7-a^2\right)\left(7+a^2\right)=7^2-\left(a^2\right)^2=49-a^4\\ \left(x-y\right)^2-\left(x+y\right)^2=\left(x-y+x+y\right)\left(x-y-x-y\right)=2x\cdot-2y=-4xy\\ g)\left(2x^3-\dfrac{1}{2}y\right)^2=\left(2x^3\right)^2-2\cdot2x^3\cdot\dfrac{1}{2}y+\left(\dfrac{1}{2}y^2\right)=4x^6-2x^3y+\dfrac{1}{4}y^2\\ h)\left(x^2+4y\right)^2=\left(x^2\right)^2+2\cdot x^2\cdot4y+\left(4y\right)^2=x^4+8x^2y+16y^2\\ i)\left(a+b+c\right)^2=\left[a+\left(b+c\right)\right]^2=a^2+2a\left(b+c\right)+\left(b+c\right)^2\\ =a^2+2ab+2ac+b^2+2bc+c^2=a^2+b^2+c^2+2ab+2bc+2ac\\ k)\left(a-b-c\right)^2=\left[a-\left(b+c\right)\right]^2=a^2-2a\left(b+c\right)+\left(b+c\right)^2\\ =a^2-2ab-2ac+b^2+2bc+c^2=a^2+b^2+c^2-2ab-2ac+2bc\)
Bài 2:
\(a)2x^2y-\dfrac{1}{4}x^2y+5x^2y-4x^2y\\ =x^2y\cdot\left(2-\dfrac{1}{4}+5-4\right)\\ =x^2y\cdot\left(3-\dfrac{1}{4}\right)\\ =\dfrac{11}{4}x^2y\\ b)5y^3z^2-3y^3z^2+7y^3z^2-6y^3z^2\\ =y^3z^2\cdot\left(5-3+7-6\right)\\ =3y^3z^2\\ c)-4x^3y^4+6x^2y^3+\dfrac{1}{2}x^3y^4-\dfrac{3}{2}x^2y^3\\ =\left(\dfrac{1}{2}x^3y^4-4x^3y^4\right)+\left(6x^2y^3-\dfrac{3}{2}x^2y^3\right)\\ =x^3y^4\left(\dfrac{1}{2}-4\right)+x^2y^3\left(6-\dfrac{3}{2}\right)\\ =-\dfrac{7}{2}x^3y^4+\dfrac{9}{2}x^2y^3\)
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: \(2\left(5-x\right)\left(5+x\right)-\left(2x+3\right)^2-x\left(3x+2\right)\)
\(=2\left(25-x^2\right)-4x^2-12x-9-3x^2-2x\)
\(=2\left(25-x^2\right)-7x^2-14x-9\)
\(=50-2x^2-7x^2-14x-9=-9x^2-14x+41\)
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 \(B=-4\cdot0-26=-26\)