Cho biết:
A = 5x + 15y + 8x + 24y
Tại x = 2; y = 10 thì biểu thức trên có giá trị bằng bao nhiêu?
K
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9 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
10 tháng 10 2021
a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)
b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)
c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Xem lại đề câu d
DG
1
13 tháng 4 2020
a/ \(5x-2y=23\)
\(\Leftrightarrow y=\frac{5x-23}{2}=\frac{6x-12-\left(x+11\right)}{2}=x-6-\frac{x+11}{2}\)
Vì x, y nguyên nên \(\frac{x+11}{2}=t\in Z\)
\(\Rightarrow\left\{{}\begin{matrix}x=2t-11\\y=t-6\end{matrix}\right.\) (t nguyên tùy ý)
Để $x,y$ nguyên dương thì \(\left\{{}\begin{matrix}x=2t-11>0\\y=t-6>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}t>\frac{11}{2}\\t>6\end{matrix}\right.\)
\(\Leftrightarrow t>6\)
Vậy nghiệm nguyên dương \(\left\{{}\begin{matrix}x=2t-11\\y=t-6\end{matrix}\right.\)\(t\in Z;t>6\)
C
29 tháng 6 2017
1) x2 + xy-8x-8y
= x.(x+y) -8.(x+y)
= (x+y).(x-8)
2) 9x2 -6x + 1 -36y2
=(3x-1)2 -(6y)2
=(3x-1-6y)(3x-1+6y)
3)a2 - b2 -12a+12b
= (a-b)(a+b)-12.(a-b)
=(a-b).(a+b-12)
4)x2 -2xy+8x-16y
=x.(x-2y)+8.(x-2y)
=(x-2y).(x+8)
5) x2 -9y2 +5x+15y
=(x-3y)(x+3y)+5.(x+3y)
= (x+3y).(x-3y+5)
TN
29 tháng 6 2017
\(1,x^2+xy-8x-8y=x\left(x+y\right)-8\left(x+y\right)=\left(x-8\right)\left(x+y\right)\)\(3,a^2-b^2-12a+12b=\left(a-b\right)\left(a+b\right)-12\left(a-b\right)=\left(a-b\right)\left(a+b-12\right)\)\(4,x^2-2xy+8x-16y=x\left(x-2y\right)+8\left(x-2y\right)=\left(x+8\right)\left(x-2y\right)\)\(5,x^2-9y^2+5x+15y=\left(x-3y\right)\left(x+3y\right)+5\left(x+3y\right)=\left(x+3y\right)\left(x-3y+5\right)\)
30 tháng 6 2021
a) Ta có: \(8x\left(2x-3\right)-4x\left(4x+3\right)=72\)
\(\Leftrightarrow16x^2-24x-16x^2-12x=72\)
\(\Leftrightarrow-36x=72\)
hay x=-2
b) Ta có: \(\left(x+2\right)\left(x+4\right)-x\left(x+2\right)=104\)
\(\Leftrightarrow x^2+6x+8-x^2-2x=104\)
\(\Leftrightarrow4x=96\)
hay x=24
c) Ta có: \(\left(x-1\right)\left(x+4\right)-x\left(x-1\right)=308\)
\(\Leftrightarrow x^2+3x-4-x^2+x=308\)
\(\Leftrightarrow4x=312\)
hay x=78
d) Ta có: \(15x\left(2x-3\right)-\left(5x+2\right)\left(6x-5\right)=-22\)
\(\Leftrightarrow30x^2-45x-30x^2+25x-12x+10=-22\)
\(\Leftrightarrow-32x=-32\)
hay x=1
5 tháng 3 2018
x2 + 15y2 + xy + 8x + y + 2016
\(=\left(x+\frac{y}{2}+4\right)^2+\frac{45}{5}\left(y-\frac{2}{5}\right)^2-535,25\ge535,25\)
\(\Rightarrow Min_A=-535,25\text{ khi }x=\frac{-61}{15};y=\frac{2}{15}\)
TT
0
ai trả lời giùm đi, dễ quá mak
Bài giải
\(A=3x+15y+8x+24y\)
\(A=x\left(3+8\right)+y\left(15+24\right)\)
\(A=x\cdot11+y\cdot39\)
Thay x = 2 , y = 10 và biểu thức A ta có :
\(A=2\cdot11+10\cdot39\)
\(A=22+390\)
\(A=412\)