A=(-5/9)x3/11+(-13/18)x3/11
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a) \(\left(\frac{1}{5}+\frac{3}{5}\right)\times\frac{1}{4}\)
\(=\frac{4}{5}\times\frac{1}{4}\)
\(=\frac{4}{20}=\frac{1}{5}\)
\(\left(\frac{1}{5}+\frac{3}{5}\right)\times\frac{1}{4}=\frac{1}{5}\times\frac{1}{4}+\frac{3}{5}\times\frac{1}{4}\)
\(=\frac{1}{20}+\frac{3}{20}\)
\(=\frac{4}{20}=\frac{1}{5}\)
b) \(\left(\frac{7}{9}-\frac{2}{9}\right)\times\frac{3}{2}\)
\(=\frac{5}{9}\times\frac{3}{2}\)
\(=\frac{15}{18}=\frac{5}{6}\)
\(\left(\frac{7}{9}-\frac{2}{9}\right)\times\frac{3}{2}=\frac{7}{9}\times\frac{3}{2}-\frac{2}{9}\times\frac{3}{2}\)
\(=\frac{21}{18}-\frac{6}{18}\)
\(=\frac{15}{18}=\frac{5}{6}\)
c) \(\left(\frac{11}{13}-\frac{9}{13}\right)\times\frac{39}{21}\)
\(=\frac{2}{13}\times\frac{39}{21}\)
\(=\frac{78}{273}=\frac{2}{7}\)
\(\left(\frac{11}{13}-\frac{9}{13}\right)\times\frac{39}{21}=\frac{11}{13}\times\frac{39}{21}-\frac{9}{13}\times\frac{39}{21}\)
\(=\frac{429}{273}-\frac{351}{273}\)
\(=\frac{78}{273}=\frac{2}{7}\)
a,\(\left(\frac{1}{5}+\frac{3}{5}\right)x\frac{1}{4}\)
=\(\frac{4}{5}x\frac{1}{4}\)
=\(\frac{4}{20}\)
=\(\frac{1}{5}\)
,\(\left(\frac{1}{5}+\frac{3}{5}\right)x\frac{1}{4}\)
=\(\frac{4}{5}x\frac{1}{4}\)
=\(\frac{4x1}{5x4}\)
=\(\frac{1x1}{5x1}\)
=\(\frac{1}{5}\)
\(a.=\left(\dfrac{4}{5}.\dfrac{5}{6}\right).\dfrac{2}{3}=\dfrac{4}{6}.\dfrac{2}{3}=\dfrac{4}{9}\)
\(b.\dfrac{4}{5}.\dfrac{3}{4}+\dfrac{5}{4}.\dfrac{3}{4}=\dfrac{3}{5}+\dfrac{15}{16}=\dfrac{123}{80}\)
\(c.\left(\dfrac{11}{23}+\dfrac{9}{23}\right)+\left(\dfrac{2}{23}+\dfrac{18}{23}\right)=\dfrac{20}{23}+\dfrac{20}{23}=\dfrac{40}{23}\)
\(d.\left(\dfrac{27}{12}-\dfrac{25}{36}\right)+\left(\dfrac{17}{6}-\dfrac{15}{6}\right)=\dfrac{14}{9}+\dfrac{1}{3}=\dfrac{17}{9}\)
xin loi nhe minh chi biet lam bai c thoi a hihihihihihhi
c) (2008 - 8) : 2 + 1 = 1001
4) Ta có: \(\dfrac{2x-5}{5}-\dfrac{x+3}{3}=\dfrac{2-3x}{2}-x-2\)
\(\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}-\dfrac{10\left(x+3\right)}{30}=\dfrac{15\left(2-3x\right)}{30}-\dfrac{30\left(x+2\right)}{30}\)
\(\Leftrightarrow12x-30-10x-30=30-45x-30x-60\)
\(\Leftrightarrow-22x-60=-75x-30\)
\(\Leftrightarrow-22x+75x=-30+60\)
\(\Leftrightarrow53x=30\)
\(\Leftrightarrow x=\dfrac{30}{53}\)
Vậy: \(S=\left\{\dfrac{30}{53}\right\}\)
5) Ta có: \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=5\)
\(\Leftrightarrow\dfrac{2\left(5x-3\right)}{12}-\dfrac{3\left(7x-1\right)}{12}=\dfrac{60}{12}\)
\(\Leftrightarrow10x-6-21x+3=60\)
\(\Leftrightarrow-11x-3=60\)
\(\Leftrightarrow-11x=63\)
\(\Leftrightarrow x=-\dfrac{63}{11}\)
Vậy: \(S=\left\{-\dfrac{63}{11}\right\}\)
`9,x^3+x^2-2=0`
`x^3-x^2+2x^2-2=0`
`<=>x^2(x-1)+2(x-1)(x+1)=0`
`<=>(x-1)(x^2+2x+2)=0`
`<=>x=1`
`14,x^2-2x+1=0`
`<=>(x-1)^2=0`
`<=>x-1=0`
`<=>x=1`
`15,x^3+3x^2+3x+1=0`
`<=>(x+1)^3=0`
`<=>x+1=0`
`<=>x=-1`
\(\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\times\dfrac{3}{7}=\dfrac{11}{11}\times\dfrac{3}{7}=1\times\dfrac{3}{7}=\dfrac{3}{7}\)
\(\dfrac{3}{5}\left(\dfrac{7}{9}-\dfrac{2}{9}\right)=\dfrac{3}{5}\times\dfrac{5}{9}=\dfrac{1}{3}\)
Bài 6:
1) Ta có: \(2x\left(x-5\right)-\left(x+3\right)^2=3x-x\left(5-x\right)\)
\(\Leftrightarrow2x^2-10x-\left(x^2+6x+9\right)=3x-5x+x^2\)
\(\Leftrightarrow2x^2-10x-x^2-6x-9-3x+5x-x^2=0\)
\(\Leftrightarrow-14x-9=0\)
\(\Leftrightarrow-14x=9\)
\(\Leftrightarrow x=-\dfrac{9}{14}\)
Vậy: \(S=\left\{-\dfrac{9}{14}\right\}\)
`1)2x(x-5)-(x+3)^2=3x-x(5-x)`
`<=>2x^2-10x-x^2-6x-9=3x-5x+x^2`
`<=>x^2-16x-9=x^2-2x`
`<=>14x=-9`
`<=>x=-9/14`
Lời giải:
a.
$x^3+y^3=(x+y)^3-3xy(x+y)=9^3-3.9.18=243$
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=[9^2-2.18]^2-2.18^2=1377$
Nếu $x\geq y$ thì:
$x^3-y^3=(x-y)(x^2+xy+y^2)$
$=|x-y|[(x+y)^2-xy]=\sqrt{(x+y)^2-4xy}[(x+y)^2-xy]$
$=\sqrt{9^2-4.18}(9^2-18)=189$
Nếu $x< y$ thì $x^3-y^3=-189$
b.
$A=(x+y)^2-6(x+y)+y-5$
$=(-9)^2-6(-9)+y-5=130+y$
Chưa đủ cơ sở để tính biểu thức.
\(3\dfrac{1}{3}.x+11\dfrac{3}{4}=\left(-13,25\right)\)
\(\dfrac{10}{3}.x+\dfrac{47}{4}=\dfrac{-53}{4}\)
\(\dfrac{10}{3}.x=\dfrac{-53}{4}-\dfrac{47}{4}\)
\(\dfrac{10}{3}.x=-25\)
\(x=-25:\dfrac{10}{3}\)
\(x=\dfrac{-15}{2}\)
Đúng mình tích cho ạ giúp vs
\(A=\left(-\dfrac{5}{9}\right).\dfrac{3}{11}+\left(-\dfrac{13}{18}\right).\dfrac{3}{11}=\dfrac{3}{11}\left(-\dfrac{5}{9}-\dfrac{13}{18}\right)\)
\(=\dfrac{3}{11}.\dfrac{-23}{18}=-\dfrac{23}{66}\)