Tính
a,(\(\frac{2}{3}\))\(^3\)
b,(-2 \(\frac{3}{4}\))\(^2\)
c,(0,6)\(^4\)
d,(-1/2)\(^4\)
e,(-1/2)\(^5\)
GIÚP MINK VỚI !
Tính
a) \(\left( {\frac{4}{5} - 1} \right):\frac{3}{5} - \frac{2}{3}.0,5\)
b) \(1 - {\left( {\frac{5}{9} - \frac{2}{3}} \right)^2}:\frac{4}{{27}}\)
c)\(\left[ {\left( {\frac{3}{8} - \frac{5}{{12}}} \right).6 + \frac{1}{3}} \right].4\)
d) \(0,8:\left\{ {0,2 - 7.\left[ {\frac{1}{6} + \left( {\frac{5}{{21}} - \frac{5}{{14}}} \right)} \right]} \right\}\)
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
giúp mik vs mai mik kiểm tra rùi
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
d)$(x+2)^{2}$ - (x-1).(x+3) = (2x-4).(x+4)-3
e)$(2x-3)^{2}$ = (2x-3).(x+1)
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
Bài 1: Tính
a) \(\frac{30x^3}{11y^2}.\frac{121y^5}{25x}\)
b) \(\frac{24y^5}{7x^2}.\frac{-21x}{12y^3}\)
c) \(\left(\frac{-18y^3}{25x^4}\right).\left(\frac{-15x^2}{9y^3}\right)\)
d) \(\frac{3x^2}{2y}.\frac{1}{4y}.\frac{5}{3y}\)
e) \(\frac{2x}{3}.\frac{x+1}{2x}\)
g) \(\frac{5-x}{x-3}.\frac{2}{3}.\frac{x}{4}\)
( giúp mink vs mink đag cần gấp)
\(a,\frac{30x^3}{11y^2}.\frac{121y^5}{25x}\)
\(=>\frac{30x^3.121y^5}{11y^2.25x}=\frac{6x^2.11y^3}{5}=\frac{66x^2.y^3}{5}\)
\(b,\frac{24y^5}{7x^2}.\frac{-21x}{12y^3}\)
\(=>\frac{24y^5.\left(-21\right)x}{7x^2.12y^3}=\frac{2y^2.\left(-3\right)}{x}=-\frac{6y^2}{x}\)
\(c,\left(\frac{-18y^3}{25x^4}\right).\left(\frac{-15x^2}{9y^3}\right)\)
\(=>\frac{-18y^3.\left(-15\right)x^2}{25x^4.9y^3}=\frac{-2.\left(-3\right)}{5x^2}=\frac{6}{5x^2}\)
\(d,\frac{3x^2}{2y}.\frac{1}{4y}.\frac{5}{3y}\)
\(=>\frac{3x^2.1.5}{2y.4y.3y}=\frac{15x^2}{24y^3}=\frac{5x^2}{8y^3}\)
\(e,\frac{2x}{3}.\frac{x+1}{2x}\)
\(=>\frac{2x\left(x+1\right)}{3.2x}=\frac{x+1}{3}\)
\(g,\frac{5-x}{x-3}.\frac{2}{3}.\frac{x}{4}\)
\(=>\frac{2x\left(5-x\right)}{3.4\left(x-3\right)}=\frac{10x-2x^2}{12\left(x-3\right)}=\frac{10x-2x^2}{12x-9}\)
Giai giup mk
a) (-2,5) . (-7,5) . (-4)
b)\(1\frac{4}{23}+\frac{8}{21}-\frac{4}{23}+0,6+\frac{13}{21}\)
c)\(\frac{2}{7}.15\frac{1}{3}-\frac{2}{7}.20\frac{1}{3}+4\frac{1}{3}\)
d)\(2\frac{1}{4}:\left(\frac{-3}{5}\right)-1\frac{1}{4}:\left(\frac{-3}{5}\right)\)
a) ( -2.5 ) . ( 7,5) .( -4 )
= [(-2,5).(-4)].(7,5)
= 10 . 7,5
= 75
b) \(1\frac{4}{23}+\frac{8}{21}-\frac{4}{23}+0,6+\frac{13}{21}\)
=\(1\frac{4}{23}-\frac{4}{23}+\frac{8}{21}+\frac{13}{21}-0,6\)
\(=1+1-0,6\)
\(=2-0,6\)
= 1,4
c) \(\frac{2}{7}.15\frac{1}{3}-\frac{2}{7}.20.\frac{1}{3}+4\frac{1}{3}\)
\(=\frac{2}{7}.5-\frac{1}{3}.\frac{40}{7}+4\frac{1}{3}\)
= \(=\frac{10}{7}-\frac{17}{7}\)
= -1
d) \(2\frac{1}{4}:\left(\frac{-3}{5}\right)-1\frac{1}{4}:\left(\frac{-3}{5}\right)\)
\(=\frac{9}{4}.\left( \frac{-5}{3}\right)-\frac{5}{4}.\left(\frac{-5}{3}\right)\)
=\(\left(\frac{-5}{3}\right).\left(\frac{9}{4}-\frac{5}{4}\right)\)
\(=\frac{-5}{3}.1\)
\(=\frac{-5}{3}\)
Tính giá trị biểu thức
\(A=\frac{\left(0,6\right)^5}{\left(0,3\right)^6}\) \(B=\frac{-1^3}{3^3+3\times6^2+6^3}\)
\(C=\frac{2\times5^5\times4^5}{20^4\times5^4}\) \(D=\left(\frac{-10}{3}\right)^3\times\left(\frac{-6}{5}\right)^3\)
\(E=2\div\left(\frac{1}{2}-\frac{2}{3}\right)^{^2}\) \(G=\left(\frac{4}{5}-\frac{3}{2}\right)^2\div\left(1-\frac{1}{4}+\frac{2}{3}\right)\)
Bài 1:
a, x+1/2=3/8.4/5
b, 5/16:x-1/4=5/8
c,-1/4.x+3/7.x=2
d, 22/9-(x+1/2)^2=7/3
e, \(\frac{1}{4}-x+5\frac{1}{8}=6\frac{1}{8}\)
Giúp mik với, đang gấp lắm!!!!!!!!!!!!!!!!!!!!!!!!!!!!
giải các phương trình
\(A.\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\) \(B.\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
\(C.\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\) \(D.\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)
mọi người ơi giúp mink vs ạ vì mink đng cần gấp
có ai bt ko giúp mink với
ai giúp mink ik ạ mink đng cần gấp