0,4x² - 4x -1200 =0

\(\Delta=b^2-4ac\) = (-4)² - 4. (1200) . 0,4 = 1936

=> pt có 2 nghiệm phân biệt.

x1= \(\dfrac{-\left(-4\right)-\sqrt{1936}}{0,4}\)= 100

x2=\(\dfrac{-\left(-4\right)+\sqrt{1936}}{0,4}\)= 120

`Answer:`

\(3\left(\frac{5}{3}x-7\right)-2\left(1.5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Leftrightarrow3\left(\frac{5x}{3}-7\right)-2\left(5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Leftrightarrow5x-21-10x-12-5x-20+x^2+4x=80+x^2\)

\(\Leftrightarrow5x-21-10x-12-5x-20+4x=80\)

\(\Leftrightarrow-6x-53=80\)

\(\Leftrightarrow-6x=133\)

\(\Leftrightarrow x=-\frac{133}{6}\)

\(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\left(\frac{1}{5}x-\frac{2}{3}\right)\left(4\frac{x^2}{3}+1\right)=\frac{22}{45}x^2\)

\(\Leftrightarrow36x^2\left(\frac{x}{3}-\frac{1}{2}\right)-45\left(\frac{x}{5}-\frac{2}{3}\right)\left(\frac{4x^2}{3}+1\right)=22x^2\)

\(\Leftrightarrow12x^3-18x^2-12x^3-9x+40x^2+30=22x^2\)

\(\Leftrightarrow22x^2-9x+30=22x^2\)

\(\Leftrightarrow-9x+30=0\)

\(\Leftrightarrow-9x=-30\)

\(\Leftrightarrow x=\frac{10}{3}\)

2: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{4x+y-z}{4\cdot3+12-16}=\dfrac{8}{8}=1\)

Do đó: x=3; y=12; z=16

Đây bạn nhé =))

\(\begin{array}{l} a)3x - 1 = x + 3\\ \Leftrightarrow 3x - x = 3 + 1\\ \Leftrightarrow 2x = 4\\ \Leftrightarrow x = 2\\ b)15 - 7x = 9 - 3x\\ \Leftrightarrow - 7x + 3x = 9 - 15\\ \Leftrightarrow - 4x = - 6\\ \Leftrightarrow x = \dfrac{3}{2}\\ c)x - 3 = 18\\ \Leftrightarrow x = 18 + 3\\ \Leftrightarrow x = 21 \end{array}\)

\(\begin{array}{l} d)2x + 1 = 15 - 5x\\ \Leftrightarrow 2x + 5x = 15 - 1\\ \Leftrightarrow 7x = 14\\ \Leftrightarrow x = 2\\ e)3x - 2 = 2x + 5\\ \Leftrightarrow 3x - 2x = 5 + 2\\ \Leftrightarrow x = 7\\ f) - 4x + 8 = 0\\ \Leftrightarrow - 4x = - 8\\ \Leftrightarrow x = 2 \end{array}\)

Anh sửa đề câu cuối.

\(\begin{array}{l} g)4x + 5 = 3x\\ \Leftrightarrow 4x - 3x = - 5\\ \Leftrightarrow x = - 5\\ h)3 + 2.25 + 2.6 = 2x + 5.0,4x\\ \Leftrightarrow 3 + 50 + 12 = 2x + 2x\\ \Leftrightarrow 47 = 4x\\ \Leftrightarrow x = \dfrac{{47}}{4}\\ i)2x{\left( {x + 2} \right)^2} - 8{x^2} = 2\left( {{x^3} - 8} \right)\\ \Leftrightarrow 2{x^3} + 8{x^2} + 8x - 8{x^2} = 2{x^3} - 16\\ \Leftrightarrow 8x = - 16\\ \Leftrightarrow x = - 2 \end{array}\)

D

D

D

a: =>3,6-x+0,5=3,5-0,75+x

=>4,1-x=x+2,75

=>-2x=-1,35

=>x=0,675

b: =>5x^2-5x+x-1=0

=>(x-1)(5x+1)=0

=>x=1 hoặc x=-1/5

c: \(\Leftrightarrow\left(\dfrac{2-x}{2008}+1\right)=\left(\dfrac{1-x}{2009}+1\right)+\left(1-\dfrac{x}{2010}\right)\)

=>\(2010-x=0\)

=>x=2010

\(0,4x^2-4x-1200=0\)

\(x^2-10x-3000=0\)

\(x^2+50x-60x-3000=0\)

\(x\left(x+50\right)-60x\left(x+50\right)=0\)

\(\left(x+50\right)\left(x-60\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-50\\x=60\end{matrix}\right.\)

PS: Ở bước hai là nhân cả hai vế với 2,5 nhé :)