a) 3/7 + 3/4 - 1/2 =33/28-1/2=19/28

b) 6/11 + 2/9 x 12 =6/11+8/3=106/33

c) 5/7 - 8/5 : 4 =5/7-2/5=11/35

a) 3/7 + 3/4 - 1/2 = \(\dfrac{12}{28}+\dfrac{21}{28}-\dfrac{14}{28}=\dfrac{19}{28}\)

b) 6/11 + 2/9 x 12 = \(\dfrac{6}{11}+\dfrac{8}{3}=\dfrac{18}{33}+\dfrac{88}{33}=\dfrac{106}{33}\)

c) 5/7 - 8/5 : 4 = \(\dfrac{5}{7}-\dfrac{2}{5}=\dfrac{25}{35}-\dfrac{14}{35}=\dfrac{11}{35}\)

3/7 + 3/4 - 1/2 = 33/28 - 1/2 = 19/28

6/11 + 2/9 x 12 = 6/11+ 8/3 = 106/33

5/7 - 8/5 : 4 = 5/7 - 2/5 = 11/35

a, 25/12

b,  25/72

a) = 3/4 + 4/3 = 25/12

b) = 3/8 - 16/9 = -101/72 * hình như hơi sai sai:"> *

a) 2 + 4/7=18/7

b)5/6-3/5=7/30

c)3/4 x 5/6=5/12

d)8/5 : 8/7=7/5

a) 2 + 4/7 = \(\dfrac{14}{7}+\dfrac{4}{7}=\dfrac{18}{7}\)

b)5/6-3/5= \(\dfrac{25}{30}-\dfrac{18}{30}=\dfrac{7}{30}\)

c)3/4 x 5/6= \(\dfrac{5}{8}\)

d)8/5 : 8/7= \(\dfrac{8}{5}x\dfrac{7}{8}=\dfrac{56}{40}=\dfrac{7}{5}\)

a,2+4/7=14/7+4/7=18/7

b,5/6-3/5=25/30-18/30=7/30

c,3/4.5/6=5/8

d,8/5:8/7=8/5.7/8=7/5

a.(x+10) /(4*x)-8* 4 -(2*x)/x+2

-(127*x-10)/(4*x)

(5/2-127*x/4)/x

Câu a

`@` `\text {Ans}`

`\downarrow`

`1,`

`a)`

`-7/25 + (-8)/25`

`= (-7 - 8)/25`

`= -15/25`

`= -3/5`

`b)`

`6/13 + (-15)/39`

`= 18/39 + (-15)/39`

`= (18 - 15)/39`

`= 3/39`

`= 1/13`

`c)`

`5/7 + 4/(-14)`

`= 10/14 + (-4)/14`

`= (10 - 4)/14`

`= 6/14`

`= 3/7`

`d)`

`-8/18 + (-15)/27`

`= -4/9 + (-5)/9`

`= (-4-5)/9`

`= -9/9 = -1`

`2,`

`a)`

`3/5 + (-7)/4`

`= 12/20 + (-35)/20`

`= (12 - 35)/20`

`=-23/20`

`b)`

`(-2) + (-5)/8`

`= (-16)/8 + (-5)/8`

`= (-16 - 5)/8`

`= -21/8`

`c)`

`1/8 + (-5)/9`

`= 9/72 + (-40)/72`

`= (9-40)/72`

`= -31/72`

`d)`

`6/13 + (-14)/39`

`= 18/39 + (-14)/39`

`= (18 - 14)/39`

`= 4/39`

`e)`

`(-18)/24 + 15/21`

`= (-3)/4 + 5/7`

`= (-21)/28 + 20/28`

`= (-21 + 20)/28`

`= -1/28`

a. \(\dfrac{-5}{4}\) x⁴ . \(\dfrac{8}{15}\) x = \(\dfrac{-40}{60}\) x⁵ = \(\dfrac{-2}{3}\) x⁵

b. -2x\(\left(\dfrac{3}{4}x^2-x+\dfrac{1}{2}\right)\) = -\(\dfrac{-3}{2}\) x³ + 2x³ - x

c. \(x\left(x-\dfrac{1}{2}\right)\) - (x - 2)(x + 3) 

= x² - \(\dfrac{1}{2}\) x - x² - 3x - 2x - 6

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

\(d,\dfrac{6}{5}-3:\dfrac{15}{4}=\dfrac{6}{5}-3\times\dfrac{4}{15}=\dfrac{6}{5}-\dfrac{4}{5}=\dfrac{2}{5}\)

\(b,\dfrac{2}{5}+\dfrac{4}{5}:4=\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{1}{4}=\dfrac{2}{5}+\dfrac{1}{5}=\dfrac{3}{5}\)

\(a,\dfrac{5}{8}+\dfrac{4}{3}=\dfrac{15}{24}+\dfrac{32}{24}=\dfrac{47}{24}\)

\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)

\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)

\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)

\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)

Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:

\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)

Tick plzz

a: Ta có: \(\left(x-2\right)\left(x+3\right)-3\left(4x-2\right)=\left(x-4\right)^2\)

\(\Leftrightarrow x^2+x-6-12x+6-x^2+8x-16=0\)

\(\Leftrightarrow-3x=16\)

hay \(x=-\dfrac{16}{3}\)

b: Ta có: \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)

\(\Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\)

\(\Leftrightarrow-14x+7+4x-6=0\)

\(\Leftrightarrow10x=1\)

hay \(x=\dfrac{1}{10}\)

c: Ta có: \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)

\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)

\(\Leftrightarrow23x+70=10x+200\)

\(\Leftrightarrow x=10\)