a) X : ( -1/3)2= 1/3
b) (4/5)5 .x =(4/5)7
c) (x+1/2)2=1/16
d) (3x+1)3= —27
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b: \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)
c: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
e: \(a^2y^2-2axby+b^2x^2\)
\(=\left(ay\right)^2-2\cdot ay\cdot bx+\left(bx\right)^2\)
\(=\left(ay-bx\right)^2\)
f: \(100-\left(3x-y\right)^2\)
\(=\left(10-3x+y\right)\left(10+3x-y\right)\)
g: \(64x^2-\left(8a+b\right)^2\)
\(=\left(8x\right)^2-\left(8a+b\right)^2\)
\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
a)\(=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)
b)\(=\dfrac{5}{9}\times\dfrac{3}{2}=\dfrac{15}{18}=\dfrac{5}{6}\)
d)\(=\left(\dfrac{12}{8}-\dfrac{3}{8}\right)\times2=\dfrac{9}{8}\times2=\dfrac{18}{8}=\dfrac{9}{4}\)
c)\(=\dfrac{4}{3}-\dfrac{5}{6}=\dfrac{8}{6}-\dfrac{5}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
a) 1 + 4/3 = 7/3
b) 5/9 : 2/3 = 5/6
c ) 4/3 -1/3 x 5/2
= 1 x 5/2
= 5/2
d) ( 3/2 - 3/8) : 1/2
= 9/8 : 1/2
= 9/4
e) 15/16 : 3/8 x 3/4
= 5/2 x 3/4
= 15/8
f) 7/19 x 1/3 x 7/19 x 2/3
= 7/19 x (1/3 x 2/3)
= 7/19 x 2/9
= 14/171
g) 3/5 x 8/27 x 25/3
= 3/5 x 25/3 x 8/27
= 5 x 8/27
= 40/27
h) 1/5 + 4/11 + 4/5 + 7/11
= (1/5 + 4/5) + (4/11 + 7/11)
= 1 + 1
= 2
a: =>31-x=60
=>x=-29
b: =>(x-140):35=280-270=10
=>x-140=350
=>x=490
c: =>(1900-2x):35=48
=>1900-2x=1680
=>2x=220
=>x=110
d: =>\(2^{2x-1}=2^9\cdot2=2^{11}\)
=>2x-1=11
=>x=6
e: =>(x+2)^5=4^5
=>x+2=4
=>x=2
f: =>3x-4=0 hoặc x-1=0
=>x=4/3 hoặc x=1
g: =>(2x-1)^2=49
=>2x-1=7 hoặc 2x-1=-7
=>x=-3 hoặc x=4
h: =>x(x+1)/2=78
=>x(x+1)=156
=>x=12
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
a: \(=\dfrac{-3}{5}\cdot\dfrac{5}{7}+\dfrac{-3}{5}\cdot\dfrac{3}{7}+\dfrac{-3}{5}\cdot\dfrac{6}{7}\)
\(=\dfrac{-3}{5}\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)=\dfrac{-3}{5}\cdot2=-\dfrac{6}{5}\)
b: \(=\dfrac{3}{13}\cdot\dfrac{6}{11}+\dfrac{3}{13}\cdot\dfrac{5}{11}-\dfrac{2}{13}=\dfrac{3}{13}-\dfrac{2}{13}=\dfrac{1}{13}\)
c: =>1/2x+1+3/8=7/16
=>1/2x=-15/16
=>x=-15/8
d: =>5/2x-1/3=1/6*(-9)/2=-9/12=-3/4
=>5/2x=-3/4+1/3=-9/12+4/12=-5/12
=>x=-1/6
a) \(\dfrac{3}{5}+x=\dfrac{4}{3}\)
\(\Leftrightarrow x=\dfrac{4}{3}-\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{20}{15}-\dfrac{9}{15}\)
\(\Leftrightarrow x=\dfrac{11}{15}\)
b) \(x+\dfrac{5}{6}=\dfrac{1}{7}\)
\(\Leftrightarrow x=\dfrac{1}{7}-\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{6}{42}-\dfrac{35}{42}\)
\(\Leftrightarrow x=-\dfrac{29}{42}\)
c) \(x\times\dfrac{3}{8}=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{4}:\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{4}\times\dfrac{8}{3}\)
\(\Leftrightarrow x=2\)
d) \(\dfrac{4}{5}\times x=\dfrac{3}{7}\)
\(\Leftrightarrow x=\dfrac{3}{7}:\dfrac{4}{5}\)
\(\Leftrightarrow x=\dfrac{3}{7}\times\dfrac{5}{4}\)
\(\Leftrightarrow x=\dfrac{15}{28}\)
a, \(\dfrac{3}{5}+x=\dfrac{4}{3}\Leftrightarrow x=\dfrac{4}{3}-\dfrac{3}{5}=\dfrac{20-9}{15}=\dfrac{11}{15}\)
b, \(x+\dfrac{5}{6}=\dfrac{1}{7}\Leftrightarrow x=\dfrac{1}{7}-\dfrac{5}{6}=\dfrac{6-35}{42}=\dfrac{-29}{42}\)
c, \(\dfrac{3x}{8}=\dfrac{3}{4}\Leftrightarrow\dfrac{3x}{8}=\dfrac{6}{8}\Rightarrow3x=6\Leftrightarrow x=2\)
d, \(\dfrac{4x}{5}=\dfrac{3}{7}\Leftrightarrow\dfrac{28x}{35}=\dfrac{15}{35}\Rightarrow x=\dfrac{15}{28}\)
\(a,x:\left(\frac{-1}{3}\right)^2=\frac{1}{3}\)
\(x:\left(\frac{1}{3}\right)^2=\frac{1}{3}\)
\(x=\frac{1}{3}\times\left(\frac{1}{3}\right)^2=\frac{1}{27}\)
\(b,\left(\frac{4}{5}\right)^5.x=\left(\frac{4}{5}\right)^7\)
\(x=\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5\)
\(x=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)
\(c,\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{-1}{4}\)
\(d,\left(3x+1\right)^3=-27\)
\(\left(3x+1\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+1=-3\)
\(3x=-4\)
\(x=\frac{-4}{3}\)