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 

\(\Rightarrow\orbr{\begin{cases}x-25=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=25\\x=-1\end{cases}}}\)

5x^2-120x-125=0

x^2-24x-25=0

x^2+x-25x-25=0

X x (x+1)-25(x+1)=0

(X+1) x (X-25)=0

x+1=0 ; x-25= 0

x1=-1

x2=25

\(x=\dfrac{\left(\dfrac{18}{60}-\dfrac{16}{60}-\dfrac{21}{60}\right)\cdot\dfrac{5}{19}}{\left(\dfrac{5}{70}+\dfrac{10}{70}+\dfrac{6}{70}\right)\cdot\dfrac{-4}{3}}\)

\(=\dfrac{\dfrac{-19}{60}\cdot\dfrac{5}{19}}{\dfrac{3}{10}\cdot\dfrac{-4}{3}}=\dfrac{-5}{60}:\dfrac{-4}{10}=\dfrac{1}{12}\cdot\dfrac{5}{2}=\dfrac{5}{24}\)

Khi x=5/24 thì \(P=\sqrt{120\cdot\dfrac{5}{24}+39}=\sqrt{25+39}=8\)

\(144x^2-120x+26=\left(144x^2-120x+25\right)+1=\left(12x-5\right)^2+1\ge0+1=1\Rightarrowđpcm\)

\(b,E=\left(9x^2-30x+25\right)+\left(16y^2+8y+1\right)=\left(3x-5\right)^2+\left(4y+1\right)^2\ge0\left(đpcm\right)\)

\(\dfrac{20x^2+120x+180}{\left(3x+5\right)^2-4x^2}+\dfrac{5x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{3\left(x^2+8x+15\right)}\)

\(=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{3\left(x+3\right)\left(x+5\right)}\)

\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\cdot\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{5\left(x+1\right)\left(x-5\right)}-\dfrac{\left(x+3\right)\cdot3\left(x+1\right)}{3\left(x+3\right)\left(x+5\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{x+5}{x+1}-\dfrac{x+1}{x+5}\)

\(=\dfrac{4\left(x+3\right)^2+\left(x+5\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x+5\right)}\)

\(=\dfrac{4x^2+24x+36+x^2+10x+25-x^2-2x-1}{\left(x+1\right)\cdot\left(x+5\right)}\)

\(=\dfrac{4x^2+32x+60}{\left(x+1\right)\left(x+5\right)}=\dfrac{4\left(x^2+8x+15\right)}{\left(x+1\right)\left(x+5\right)}\)

\(=\dfrac{4\left(x+3\right)\cdot\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}=\dfrac{4x+12}{x+1}\)

120x5/6=100

5/6x76=190/3

120 x 5/6

= 600/6 = 100/1 = 100

5/6 x 76

= 380/6 = 190/3

Cảm ơn nha

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''

a, ( 314 - x ) - 42 = 0

    ( 314 - x ) = 0 + 42

    ( 314 - x ) =   42

              x   = 314 - 42 

              x   =    272

b, 540 + ( 345 - x ) = 740

              ( 345 - x ) = 740 - 540

              ( 345 - x ) =  200

                        x   =  345 - 200

                        x   =    145

c, ( x - 72 ) : 36 = 418

    ( x - 72 ) = 418 x 36

    ( x - 72 ) =    15 048

     x          =  15 048 + 72

     x         =      15 120

k mk nha

Khi x=100 thì \(C=\dfrac{0.0002\cdot100^2+120\cdot100+1000}{100}=\dfrac{6501}{50}\)

Khi x=1000 thì \(C=\dfrac{0.0002\cdot1000^2+120\cdot1000+1000}{1000}=\dfrac{606}{5}\)

\(\left(84,6-2\cdot x\right):3,02=5,1\)

\(\Rightarrow84,6-2\cdot x=15,402\)

\(\Rightarrow2\cdot x=69,198\)
\(\Rightarrow x=69,198:2\)

\(\Rightarrow x=34,599\)

_____________

\(\left(15\cdot24-x\right):0,25=100:0,25\)

\(\Rightarrow\left(360-x\right):0,25=400\)

\(\Rightarrow360-x=100\)

\(\Rightarrow x-360-100\)

\(\Rightarrow x=260\)

______________

\(128\cdot x-12\cdot x-16\cdot x=5200\)

\(\Rightarrow x\cdot\left(128-12-16\right)=5200\)

\(\Rightarrow x\cdot100=5200\)

\(\Rightarrow x=5200:100\)

\(\Rightarrow x=52\)

__________________

\(5\cdot x+3,75\cdot x+1,25\cdot x=20\)

\(\Rightarrow x\cdot\left(5+3,75+1,25\right)=20\)

\(\Rightarrow10\cdot x=20\)

\(\Rightarrow x=20:10\)

\(\Rightarrow x=2\)

\(x\cdot3,7+x\cdot6,3=360:120\)

\(\Rightarrow x\cdot\left(3,7+6,3\right)=3\)

\(\Rightarrow x\cdot10=3\)

\(\Rightarrow x=\dfrac{3}{10}\)

__________________

\(x\cdot23-6\cdot23+x\cdot69=320\)

\(\Rightarrow x\cdot\left(23+69\right)=320+6\cdot23\)

\(\Rightarrow x\cdot92=458\)

\(\Rightarrow x=458:92\)

\(\Rightarrow x=\dfrac{229}{46}\)

___________________

\(\left(x+1\right)\left(x+2\right)=72\)

\(\Rightarrow x^2+2x+x+2=72\)

\(\Rightarrow x^2+3x+2=72\)

\(\Rightarrow x^2+3x+2-72=0\)

\(\Rightarrow x^2+3x-70=0\)

\(\Rightarrow x^2+10x-7x-70=0\)

\(\Rightarrow\left(x-7\right)\left(x+10\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-10\end{matrix}\right.\)

___________________

\(\left(x+2\right)\cdot16\cdot x=160x\)

\(\Rightarrow16x^2+32x=160x\)

\(\Rightarrow16x^2+32x-160x=0\)

\(\Rightarrow16x^2-128x=0\)

\(\Rightarrow16x\left(x-8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)

Đổi:

f.x.23-6.23+x.69=230

=>23.(x-6+3x)=230

=>4x-6=10

=>4x=10+6

=>4x=16

=>x=4

h.(x+2).16.x=160.x

=>(16x+32).x = 160x

=>16x^2 + 32x = 160x

=>16x^2 = 128x

=>16x^2 - 128x=0

=>x(16x-128)=0

=>x=0 hoặc 16x-128=0

+)x=0=>x=0

+)16x-128=0

=>16x=128

=>x=8

\(A=x^2+3x-5=x^2+3x+\frac{9}{4}-\frac{29}{4}\)

\(=\left(x+\frac{3}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)

Vậy \(A_{min}=-\frac{29}{4}\Leftrightarrow x+\frac{3}{2}=0\Leftrightarrow x=-\frac{3}{2}\)