a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)

\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)

\(\Leftrightarrow2x=-8\)

hay x=-4

b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)

\(\Leftrightarrow-10x=-10\)

hay x=1

c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)

\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)

\(\Leftrightarrow-4x=-8\)

hay x=2

a: x=2/3-4/5=10/15-12/15=-2/15

b: 1/2-x=7/12

=>x=1/2-7/12=-1/12

c: =>7/2:x=-7/2

=>x=-1

d: =>1/6x=3/8-5/2=3/8-20/8=-17/8

=>x=-17/8*6=-102/8=-51/4

e: =>1,5x=-1,5

=>x=-1

a) Ta có: \(\left(2x-5\right)^3=216\)

\(\Leftrightarrow2x-5=6\)

\(\Leftrightarrow2x=11\)

hay \(x=\dfrac{11}{2}\)

b) Ta có: \(2x-3⋮x+4\)

\(\Leftrightarrow-11⋮x+4\)

\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)

hay \(x\in\left\{-3;-5;7;-15\right\}\)

Alo, sugeni two wai phem. Si ga no, you woo be the me that nas te, ai gi da

a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)

b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)

c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)

d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)

\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)

a: \(\Leftrightarrow x^2+10x+25-x^2+4x=55\)

=>14x=30

hay x=15/7

b: \(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\)

hay \(x\in\left\{7;3\right\}\)

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

3: Số học sinh giỏi là 40*1/5=8 bạn

Số học sinh trung bình là 32*3/8=12 bạn

Số học sinh khá là 32-12=20 bạn

1:

a: -1/3+7/6=7/6-2/6=5/6

b: 5/7-3/5=25/35-21/35=4/35

c: 0,75*4/5=4/5*3/4=3/5

a) Ta có: \(\left|4-5x\right|=24\)

\(\Leftrightarrow\left[{}\begin{matrix}-5x+4=24\\-5x+4=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x=20\\-5x=-28\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{-4;\dfrac{28}{5}\right\}\)

b) Ta có: \(\left(8+x\right)\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

Vậy: \(x\in\left\{-8;6\right\}\)

a) \(\left|4-5x\right|=24\)

\(\Leftrightarrow\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)

b) \(\left(8+x\right)\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

a.|4-5.x|=24

TH1: 4-5.x=24

        ⇔5x=-20

        ⇔x=-4

TH1: 4-5.x=-24

        ⇔5x=28

        ⇔x=28/5

b.(8+x).(6-x)=0

\(\left\{{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

Bài 4: 

a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)

\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)

\(1,25-x=\dfrac{11}{12}\)

\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)

\(x=\dfrac{1}{3}\)

b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)

\(x-\dfrac{7}{6}=\dfrac{13}{12}\)

\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)

\(x=\dfrac{27}{12}=\dfrac{9}{4}\)

c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)

\(4-\left(2x+1\right)=\dfrac{8}{3}\)

\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)

\(2x+1=\dfrac{20}{3}\)

\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)

\(2x=\dfrac{17}{3}\)

\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)

Bài 15:

a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)

\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)

\(=>x=\left(\dfrac{-2}{3}\right)^8\)

b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)

\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)

\(=>x=\left(\dfrac{4}{9}\right)^9\)

c) \(\left(x+4\right)^3=-125\)

\(\left(x+4\right)^3=\left(-5\right)^3\)

\(=>x+4=-5\)

\(x=-5-4\)

\(=>x=-9\)

d) \(\left(10-5x\right)^3=64\)

\(\left(10-5x\right)^3=4^3\)

\(=>10-5x=4\)

\(5x=10-4\)

\(5x=6\)

\(=>x=\dfrac{6}{5}\)

e) \(\left(4x+5\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)

Bài 16:

a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)

\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)

\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)

b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)

\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)

\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)

c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)

\(=\dfrac{7}{4}.\dfrac{-12}{5}\)

\(=\dfrac{-21}{5}\)

\(#Wendy.Dang\)

Uh, chừa sau k dám học muộn nx

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