x2+y2−13(x−y)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\left(x-y\right)\left(x+y\right)\)
\(=74\cdot100=7400\)
c: \(=\left(x+2\right)^3\)
\(=10^3=1000\)
a) \(=\left(x-y\right)\left(x+y\right)\)
Thay \(x=87;y=13\) ta đc: \(\left(87-13\right)\left(87+13\right)=74\cdot100=7400\)
b)\(=\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
Thay \(x=10;y=-1\) ta đc:
\(10^3-\left(-1\right)^3=1000-1=999\)
c)\(=\left(x+2\right)^3\)
Thay \(x=8\) ta đc: \(\left(8+2\right)^3=10^3=1000\)
d)\(=x^2-8x+16+1=\left(x-4\right)^2+1\)
Thay \(x=104\) ta đc: \(\left(104-4\right)^2+1=100^2+1=10001\)
Bài 13:
a) \(501^2\)
\(=\left(500+1\right)^2\)
\(=500^2+2\cdot500\cdot1+1^2\)
\(=250000+1000+1\)
\(=251001\)
b) \(88^2+24\cdot88+12^2\)
\(=88^2+2\cdot12\cdot88+12^2\)
\(=\left(88+12\right)^2\)
\(=100^2\)
\(=10000\)
c) \(52\cdot48\)
\(=\left(50+2\right)\left(50-2\right)\)
\(=50^2-2^2\)
\(=2500-4\)
\(=2496\)
Bài 14:
a) \(P=\left(2x-1\right)\left(4x^2+2x+1\right)+\left(x+1\right)\left(x^2-x+1\right)\)
\(P=\left(2x\right)^3-1+x^3+1\)
\(P=8x^3+x^3\)
\(P=9x^3\)
b) \(Q=\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x+y\right)\left(x^2-xy+y^2\right)+2y^3\)
\(Q=x^3-y^3-x^3-y^3+2y^3\)
\(Q=-2y^3+2y^3\)
\(Q=0\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2+y^2}{4+9}=\dfrac{13}{13}=1\\ \Rightarrow\left\{{}\begin{matrix}x^2=4\\y^2=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm2\\y=\pm3\end{matrix}\right.\)
Vì x và y là hai đại lượng tỉ lệ thuận nên x 1 x 2 = y 1 y 2 hay
y 1 1 3 = 12 1 6 ⇒ y 1 = 24
Đáp án cần chọn là A
\(C=x^2+y^2+4x-6y+1\)
\(=x^2+4x+4+y^2-6y+9-12\)
\(=\left(x+2\right)^2+\left(y-3\right)^2+1\)
\(=30^2+10^2+1\)
=1001
6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)
\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)
7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)
\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)
8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)
\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)
9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)
10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)
\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)
11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)
12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)
13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)
Ta có: x 2 – y 2 = (x + y)(x – y)
Thay x = 87, y = 13, ta được:
x 2 – y 2 = (x + y)(x – y)
= (87 + 13)(87 – 13)
= 100.74 = 7400
mk copy trên trang này
https://lazi.vn/edu/exercise/311935/cho-cac-so-thoa-man-2x-3y-13-tim-gia-tri-nho-nhat-cua-q
\(2x+3y=13\Rightarrow y=\dfrac{13-2x}{3}\)
\(Q=x^2+\left(\dfrac{13-2x}{3}\right)^2=\dfrac{13}{9}x^2-\dfrac{52}{9}x+\dfrac{169}{9}\)
\(Q=\dfrac{13}{9}\left(x-2\right)^2+13\ge13\)
\(Q_{min}=13\) khi \(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Ta có: \(x^2-y^2-13\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)-13\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-13\right)\)