120x -20x =1600
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21 tháng 1
\(\dfrac{20x^2+120x+180}{\left(3x+5\right)^2-4x^2}+\dfrac{5x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{3\left(x^2+8x+15\right)}\)
\(=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\cdot\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{5\left(x+1\right)\left(x-5\right)}-\dfrac{\left(x+3\right)\cdot3\left(x+1\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{x+5}{x+1}-\dfrac{x+1}{x+5}\)
\(=\dfrac{4\left(x+3\right)^2+\left(x+5\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4x^2+24x+36+x^2+10x+25-x^2-2x-1}{\left(x+1\right)\cdot\left(x+5\right)}\)
\(=\dfrac{4x^2+32x+60}{\left(x+1\right)\left(x+5\right)}=\dfrac{4\left(x^2+8x+15\right)}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)\cdot\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}=\dfrac{4x+12}{x+1}\)
KN
23 tháng 11 2019
\(A=x^2+3x-5=x^2+3x+\frac{9}{4}-\frac{29}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)
Vậy \(A_{min}=-\frac{29}{4}\Leftrightarrow x+\frac{3}{2}=0\Leftrightarrow x=-\frac{3}{2}\)
H9
HT.Phong (9A5)
CTVHS
19 tháng 8 2023
\(\left(84,6-2\cdot x\right):3,02=5,1\)
\(\Rightarrow84,6-2\cdot x=15,402\)
\(\Rightarrow2\cdot x=69,198\)
\(\Rightarrow x=69,198:2\)
\(\Rightarrow x=34,599\)
_____________
\(\left(15\cdot24-x\right):0,25=100:0,25\)
\(\Rightarrow\left(360-x\right):0,25=400\)
\(\Rightarrow360-x=100\)
\(\Rightarrow x-360-100\)
\(\Rightarrow x=260\)
______________
\(128\cdot x-12\cdot x-16\cdot x=5200\)
\(\Rightarrow x\cdot\left(128-12-16\right)=5200\)
\(\Rightarrow x\cdot100=5200\)
\(\Rightarrow x=5200:100\)
\(\Rightarrow x=52\)
__________________
\(5\cdot x+3,75\cdot x+1,25\cdot x=20\)
\(\Rightarrow x\cdot\left(5+3,75+1,25\right)=20\)
\(\Rightarrow10\cdot x=20\)
\(\Rightarrow x=20:10\)
\(\Rightarrow x=2\)
H9
HT.Phong (9A5)
CTVHS
19 tháng 8 2023
\(x\cdot3,7+x\cdot6,3=360:120\)
\(\Rightarrow x\cdot\left(3,7+6,3\right)=3\)
\(\Rightarrow x\cdot10=3\)
\(\Rightarrow x=\dfrac{3}{10}\)
__________________
\(x\cdot23-6\cdot23+x\cdot69=320\)
\(\Rightarrow x\cdot\left(23+69\right)=320+6\cdot23\)
\(\Rightarrow x\cdot92=458\)
\(\Rightarrow x=458:92\)
\(\Rightarrow x=\dfrac{229}{46}\)
___________________
\(\left(x+1\right)\left(x+2\right)=72\)
\(\Rightarrow x^2+2x+x+2=72\)
\(\Rightarrow x^2+3x+2=72\)
\(\Rightarrow x^2+3x+2-72=0\)
\(\Rightarrow x^2+3x-70=0\)
\(\Rightarrow x^2+10x-7x-70=0\)
\(\Rightarrow\left(x-7\right)\left(x+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-10\end{matrix}\right.\)
___________________
\(\left(x+2\right)\cdot16\cdot x=160x\)
\(\Rightarrow16x^2+32x=160x\)
\(\Rightarrow16x^2+32x-160x=0\)
\(\Rightarrow16x^2-128x=0\)
\(\Rightarrow16x\left(x-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
27 tháng 11 2022
\(P=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3+x\right)\left(2x+3-x\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(x-5\right)\cdot5\left(x+1\right)}-\dfrac{3\left(x+1\right)\left(x+3\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{5\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{\left(x+5\right)}{x+1}-\dfrac{x+1}{x+5}\)
\(=\dfrac{5x^2+30x+45+x^2+10x+25-x^2-2x-1}{\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{5x^2+38x+69}{\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{5x^2+38x+69}{x^2+6x+5}\)
Để P là số nguyên thì \(5x^2+30x+25+8x+34⋮x^2+6x+5\)
=>\(8x+34⋮x^2+6x+5\)
=>\(\left\{{}\begin{matrix}8x+34⋮x+1\\8x+34⋮x+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x+8+26⋮x+1\\8x+40-6⋮x+5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+1\in\left\{1;-1;2;-2;13;-13;26;-26\right\}\\x+5\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\end{matrix}\right.\)
=>\(x\in\left\{-2;1\right\}\)
T
3 tháng 4 2019
1/Dùng hệ số bất định: (ko chắc nha,mình mới lớp 7)
Gọi đa thức trên là Q(x).
Thu gọn đa thức lại,ta được: \(Q\left(x\right)=x^4+30x^3+200x^2+1440x+2304\)
Giả sử \(Q\left(x\right)=\left(x^2+ax+48\right)\left(x^2+bx+48\right)\)
\(=x^4+bx^3+48x^2+ax^3+abx^2+48ax+48x^2+48bx+2304\)
Thu gọn lại,ta được: \(Q\left(x\right)=x^4+\left(a+b\right)x^3+\left(ab+96\right)x^2+\left(48a+48b\right)x+2304\)
Đồng nhất hệ số hai vế: \(\hept{\begin{cases}a+b=30\\ab+96=200\\48\left(a+b\right)=1440\end{cases}}\)
Từ a + b = 30 suy ra a = 30 - b.
Suy ra \(ab+96=b\left(30-b\right)+96=200\Rightarrow b=4\)
Suy ra a = 26.
Suy ra \(Q\left(x\right)=\left(x^2+26x+48\right)\left(x^2+4x+48\right)\)
\(=\left(x+2\right)\left(x+24\right)\left(x^2+4x+48\right)\)
DN
3
10 tháng 3 2022
120 x 5/6
= 600/6 = 100/1 = 100
5/6 x 76
= 380/6 = 190/3
= ( 120-20).x =1600
100 .x =1600
x =1600 : 100
x = 16
CHỌN CỦA MÌNH NHÉ
120x - 20x = 1600
( 120 - 20 ) x = 1600
100x = 1600
x = 1600 : 100
x = 16
Vậy x = 16