GIÚP MÌNH VỚI Ạ!
a) 4/9 + 4/3x = 7/9
b) (5/2 - x).(-4/7) = 9/14
c) 3x + 3/4 = 2\(\frac{2}{3}\)
d) -5/6 - x = 7/12 + -1/3
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a: =>4/3x=7/9-4/9=1/3
=>x=1/4
b: =>5/2-x=9/14:(-4/7)=-9/8
=>x=5/2+9/8=29/8
c: =>3x+3/4=8/3
=>3x=23/12
hay x=23/36
d: =>-5/6-x=7/12-4/12=3/12=1/4
=>x=-5/6-1/4=-10/12-3/12=-13/12
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
Bài 2
b, `\sqrt{3x^2}=x+2` ĐKXĐ : `x>=0`
`=>(\sqrt{3x^2})^2=(x+2)^2`
`=>3x^2=x^2+4x+4`
`=>3x^2-x^2-4x-4=0`
`=>2x^2-4x-4=0`
`=>x^2-2x-2=0`
`=>(x^2-2x+1)-3=0`
`=>(x-1)^2=3`
`=>(x-1)^2=(\pm \sqrt{3})^2`
`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
a) \(\left(9^4.8+9^4.5\right):\left(9^2.\left(10-1\right)\right)\)
=\(9^4.13:9^3=13.9=117\)
b) 100-(75-25)=100-50=50
a) => 4/3x = 7/9 - 4/9 = 1/3
=> x = 1/3 : 4/3 = 1/4
b) => 5/2 - x = 9/14 : (-4/7) = -9/8
=> x = 5/2 - (-9/8) = 5/2 + 9/8 = 29/8
c) => 3x = 2 và 2/3 - 3/4 = 8/3 - 3/4 = 23/12
=> x = 23/12 : 3 = 23/36
D) => -5/6 - x = 1/4
=> x = -5/6 - 1/4 = -13/12
a) \(\dfrac{4}{9}+\dfrac{4}{3}x=\dfrac{7}{9}\)
\(\dfrac{4}{3}x=\dfrac{7}{9}-\dfrac{4}{9}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:\dfrac{4}{3}\)
\(x=\dfrac{1}{4}\)
b) \(\left(\dfrac{5}{2}-x\right)\left(-\dfrac{4}{7}\right)=\dfrac{9}{14}\)
\(\dfrac{5}{2}-x=\dfrac{9}{14}:\left(-\dfrac{4}{7}\right)=-\dfrac{9}{8}\)
\(x=\dfrac{5}{2}-\left(-\dfrac{9}{8}\right)\)
\(x=\dfrac{29}{8}\)
c) \(3x+\dfrac{3}{4}=2\dfrac{2}{3}\)
\(3x+\dfrac{3}{4}=\dfrac{8}{3}\)
\(3x=\dfrac{8}{3}-\dfrac{3}{4}=\dfrac{23}{12}\)
\(x=\dfrac{23}{12}:3\)
\(x=\dfrac{23}{36}\)
d) \(-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(-\dfrac{5}{6}-x=\dfrac{1}{4}\)
\(x=-\dfrac{5}{6}-\dfrac{1}{4}\)
\(x=-\dfrac{13}{12}\)