x + 25 = 120
x = ?
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5x2-120x-125=0
x(5x-120)=0+125
x(5x-120)=125
\(\Rightarrow\orbr{\begin{cases}x=125\\5x-120=125\Rightarrow5x=125-120\Rightarrow5x=5\Rightarrow x=5:5=1\end{cases}}\)
Vậy x=125 hoặc x=1
\(5x^2-120x-125=0\)
\(5\left(x^2-24x-25\right)=0\)
\(5\left(x-25\right)\left(x+1\right)=0\)
\(\Rightarrow5=0\)vô nghiệm
\(\Rightarrow\orbr{\begin{cases}x-25=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=25\\x=-1\end{cases}}}\)
\(\dfrac{20x^2+120x+180}{\left(3x+5\right)^2-4x^2}+\dfrac{5x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{3\left(x^2+8x+15\right)}\)
\(=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\cdot\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{5\left(x+1\right)\left(x-5\right)}-\dfrac{\left(x+3\right)\cdot3\left(x+1\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{x+5}{x+1}-\dfrac{x+1}{x+5}\)
\(=\dfrac{4\left(x+3\right)^2+\left(x+5\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4x^2+24x+36+x^2+10x+25-x^2-2x-1}{\left(x+1\right)\cdot\left(x+5\right)}\)
\(=\dfrac{4x^2+32x+60}{\left(x+1\right)\left(x+5\right)}=\dfrac{4\left(x^2+8x+15\right)}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)\cdot\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}=\dfrac{4x+12}{x+1}\)
120 x 5/6
= 600/6 = 100/1 = 100
5/6 x 76
= 380/6 = 190/3
a) \(\left(314-x\right)-42=0\)
\(314-x=42\)
\(x=272\)
vay \(x=272\)
b) \(540+\left(345-x\right)=740\)
\(345-x=200\)
\(x=145\)
vay \(x=145\)
c) \(\left(x-72\right):36=418\)
\(x-72=15048\)
\(x=15120\)
vay \(x=15120\)
d) \(2010\left(x-14\right)=0\)
\(x-14=0\)
\(x=14\)
vay \(x=14\)
e) \(\left(x-2\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
vay \(\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
f) \(114-\left(x-47\right)=0\)
\(x-47=114\)
\(x=161\)
vay \(x=161\)
g) \(7272:\left(120x-91\right)=12\)
\(120x-91=606\)
\(120x=697\)
\(x=\frac{697}{120}\)
vay \(x=\frac{697}{120}\)
Despacito đã có dấu'' x= '' thì ko cần kết luận là '' vậy''
Khi x=100 thì \(C=\dfrac{0.0002\cdot100^2+120\cdot100+1000}{100}=\dfrac{6501}{50}\)
Khi x=1000 thì \(C=\dfrac{0.0002\cdot1000^2+120\cdot1000+1000}{1000}=\dfrac{606}{5}\)
\(\left(84,6-2\cdot x\right):3,02=5,1\)
\(\Rightarrow84,6-2\cdot x=15,402\)
\(\Rightarrow2\cdot x=69,198\)
\(\Rightarrow x=69,198:2\)
\(\Rightarrow x=34,599\)
_____________
\(\left(15\cdot24-x\right):0,25=100:0,25\)
\(\Rightarrow\left(360-x\right):0,25=400\)
\(\Rightarrow360-x=100\)
\(\Rightarrow x-360-100\)
\(\Rightarrow x=260\)
______________
\(128\cdot x-12\cdot x-16\cdot x=5200\)
\(\Rightarrow x\cdot\left(128-12-16\right)=5200\)
\(\Rightarrow x\cdot100=5200\)
\(\Rightarrow x=5200:100\)
\(\Rightarrow x=52\)
__________________
\(5\cdot x+3,75\cdot x+1,25\cdot x=20\)
\(\Rightarrow x\cdot\left(5+3,75+1,25\right)=20\)
\(\Rightarrow10\cdot x=20\)
\(\Rightarrow x=20:10\)
\(\Rightarrow x=2\)
\(x\cdot3,7+x\cdot6,3=360:120\)
\(\Rightarrow x\cdot\left(3,7+6,3\right)=3\)
\(\Rightarrow x\cdot10=3\)
\(\Rightarrow x=\dfrac{3}{10}\)
__________________
\(x\cdot23-6\cdot23+x\cdot69=320\)
\(\Rightarrow x\cdot\left(23+69\right)=320+6\cdot23\)
\(\Rightarrow x\cdot92=458\)
\(\Rightarrow x=458:92\)
\(\Rightarrow x=\dfrac{229}{46}\)
___________________
\(\left(x+1\right)\left(x+2\right)=72\)
\(\Rightarrow x^2+2x+x+2=72\)
\(\Rightarrow x^2+3x+2=72\)
\(\Rightarrow x^2+3x+2-72=0\)
\(\Rightarrow x^2+3x-70=0\)
\(\Rightarrow x^2+10x-7x-70=0\)
\(\Rightarrow\left(x-7\right)\left(x+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-10\end{matrix}\right.\)
___________________
\(\left(x+2\right)\cdot16\cdot x=160x\)
\(\Rightarrow16x^2+32x=160x\)
\(\Rightarrow16x^2+32x-160x=0\)
\(\Rightarrow16x^2-128x=0\)
\(\Rightarrow16x\left(x-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
\(A=x^2+3x-5=x^2+3x+\frac{9}{4}-\frac{29}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)
Vậy \(A_{min}=-\frac{29}{4}\Leftrightarrow x+\frac{3}{2}=0\Leftrightarrow x=-\frac{3}{2}\)
Giải:
\(9-3\times\left(x-9\right)=6\)
\(3\times\left(x-9\right)=9-6\)
\(3\times\left(x-9\right)=3\)
\(x-9=3:3\)
\(x-9=1\)
\(x=1+9\)
\(x=10\)
\(4+6\times\left(x+1\right)=70\)
\(6\times\left(x+1\right)=70-4\)
\(6\times\left(x+1\right)=66\)
\(x+1=66:6\)
\(x+1=11\)
\(x=11-1\)
\(x=10\)
\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\)
\(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\)
\(\dfrac{x}{13}=\dfrac{4}{13}\)
\(\Rightarrow x=4\)
\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{3}{7}\)
\(\Rightarrow x=7\)
\(5\times\left(3+7\times x\right)=40\)
\(3+7\times x=40:5\)
\(3+7\times x=8\)
\(7\times x=8-3\)
\(7\times x=5\)
\(x=5:7\)
\(x=\dfrac{5}{7}\)
\(x\times6+12:3=120\)
\(x\times6+4=120\)
\(x\times6=120-4\)
\(x\times6=116\)
\(x=116:6\)
\(x=\dfrac{58}{3}\)
\(x\times3,7+x\times6,3=120\)
\(x\times\left(3,7+6,3\right)=120\)
\(x\times10=120\)
\(x=120:10\)
\(x=12\)
\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\)
\(\left(360-x\right):0,25=400\)
\(360-x=400.0,25\)
\(360-x=100\)
\(x=360-100\)
\(x=260\)
\(71+65\times4=\dfrac{x+140}{x}+260\)
\(\left(x+140\right):x+260=71+260\)
\(x:x+140:x+260=331\)
\(1+140:x+260=331\)
\(140:x=331-1-260\)
\(140:x=70\)
\(x=140:70\)
\(x=2\)
\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\)
\(10\times x+\left(1+4+7+...+28\right)=155\)
Số số hạng \(\left(1+4+7+...+28\right)\) :
\(\left(28-1\right):3+1=10\)
Tổng dãy \(\left(1+4+7+...+28\right)\) :
\(\left(1+28\right).10:2=145\)
\(\Rightarrow10\times x+145=155\)
\(10\times x=155-145\)
\(10\times x=10\)
\(x=10:10\)
\(x=1\)
Đều theo cách lớp 5 nha em!
x=95 nhé
95 UvU