1.

=3/5x(3/7+4/7)+2/5x(13/9-4/9)

=3/5x1+2/5x1

=3/5+2/5

=1

2.Xx(3/4+4/5)=7/10

Xx31/20=7/10

X         =7/10:31/20

X         =14/31

\(\frac{3}{5}\cdot\frac{3}{7}+\frac{3}{5}\cdot\frac{4}{7}+\frac{2}{5}\cdot\frac{13}{9}-\frac{2}{5}\cdot\frac{4}{9}\)

\(=\frac{3}{5}\cdot\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{2}{5}\cdot\left(\frac{13}{9}-\frac{4}{9}\right)\)

\(=\frac{3}{5}\cdot1+\frac{2}{5}\cdot1\)\(=\frac{3}{5}+\frac{2}{5}=1\)

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\(\frac{3}{4}\cdot x+\frac{4}{5}\cdot x=\frac{7}{10}\)

\(\left(\frac{3}{4}+\frac{4}{5}\right)\cdot x=\frac{7}{10}\)

\(\frac{31}{20}\cdot x=\frac{7}{10}\)

\(x=\frac{7}{10}:\frac{31}{20}\)

\(x=\frac{14}{31}\)

Mn giúp mh với đg cần gấp lắm ln🥺🥺

\(\left(x-y\right)^7:\left(y-x\right)^6\left(dk:x\ne y\right)\)

\(=\left(x-y\right)^7:\left(x-y\right)^6\)

\(=x-y\)

a)\(\left(\frac{3}{4}+\frac{-7}{2}\right).\left(\frac{10}{11}+\frac{2}{22}\right)\)

\(\left(\frac{3}{4}+\frac{-14}{4}\right).\left(\frac{20}{22}+\frac{2}{22}\right)\)

\(\frac{-11}{4}.1=\frac{-11}{4}\)

câu b mk hiểu đề

0,75+7/12 chứ không phải = đâu

a) = 1.1204
b) = 0.5714

a) = 1.1204
b) = 0.5714

\(a,=3^{30}:3^{28}-7=3^2-7=9-7=2\\ b,=44+6\cdot10^2=44+6\cdot100=44+600=644\\ c,=178204+29478-200900=6782\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

\(=3\left(2x+3y\right)\left(x-y\right)\)

Bài 4:

$(x-1)(x^2+x+1)-x^3-6x=11$

Dùng $(x-1)(x^2+x+1)=x^3-1$:

$x^3-1-x^3-6x=11$

$-6x-1=11$

$-6x=12$

$x=-2$

Vậy $x=-2$.

$16x^2-(3x-4)^2=0$

$(4x)^2-(3x-4)^2=0$

$(4x-3x+4)(4x+3x-4)=0$

$(x+4)(7x-4)=0$

$x=-4$ hoặc $x=\dfrac47$

Vậy $x=-4,\dfrac47$.

$x^3-x^2+3-3x=0$

$=x^2(x-1)-3(x-1)=0$

$=(x-1)(x^2-3)=0$

$x-1=0$ hoặc $x^2-3=0$

$x=1$ hoặc $x=\pm\sqrt3$

Vậy $x=1,\sqrt3,-\sqrt3$.

$\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}$

Điều kiện: $x\ne-2,-1$.

$(x-1)(x+1)=(x+2)^2$

$x^2-1=x^2+4x+4$

$-4x=5$

$x=-\dfrac54$

Vậy $x=-\dfrac54$.

$\dfrac1{x+2}+\dfrac2{x+1}=0$

Điều kiện: $x\ne-2,-1$.

$\dfrac{x+1+2(x+2)}{(x+2)(x+1)}=0$

$x+1+2x+4=0$

$3x+5=0$

$x=-\dfrac53$

Vậy $x=-\dfrac53$.

$\dfrac{9-x^2}{x}:(x-3)=1$

Điều kiện: $x\ne0,3$.

$\dfrac{9-x^2}{x(x-3)}=1$

$9-x^2=x(x-3)$

$9-x^2=x^2-3x$

$2x^2-3x-9=0$

$(2x+3)(x-3)=0$

$x=-\dfrac32$ hoặc $x=3$

Nhưng $x=3$ không thỏa điều kiện.

Vậy $x=-\dfrac32$.

a)

\(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)

\(=\dfrac{146}{13}-\left(\dfrac{18}{7}+\dfrac{68}{13}\right)\)

\(=\dfrac{146}{13}-\dfrac{18}{7}-\dfrac{68}{13}\)

\(=\left(\dfrac{146}{13}-\dfrac{68}{13}\right)-\dfrac{18}{7}\)

\(=6-\dfrac{18}{7}\)

\(=\dfrac{24}{7}\)

b)

\(\dfrac{2}{7}\times5\dfrac{1}{4}-\dfrac{2}{7}\times3\dfrac{1}{4}\)

\(=\dfrac{2}{7}\times\dfrac{21}{4}-\dfrac{2}{7}\times\dfrac{13}{4}\)

\(=\dfrac{2}{7}\times\left(\dfrac{21}{4}-\dfrac{13}{4}\right)\)

\(=\dfrac{2}{7}\times2\)

\(=\dfrac{4}{7}\)

\(a,11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)

\(=\dfrac{146}{13}-\left(\dfrac{18}{7}+\dfrac{68}{13}\right)\)

\(=\dfrac{146}{13}-\dfrac{68}{13}-\dfrac{18}{7}\)

\(=6-\dfrac{18}{7}\)

\(=\dfrac{24}{7}\)

\(b,\dfrac{2}{7}\times5\dfrac{1}{4}-\dfrac{2}{7}\times3\dfrac{1}{4}\)

\(=\dfrac{2}{7}\times\dfrac{21}{4}-\dfrac{2}{7}\times\dfrac{13}{4}\)

\(=\dfrac{2}{7}\times\left(\dfrac{21}{4}-\dfrac{13}{4}\right)\)

\(=\dfrac{2}{7}\times2\)

\(=\dfrac{4}{7}\)

a) \(\left(5x-4y^3\right)\left(2x^2-1+y\right)=10x^3-5x+5xy-8x^2y^3+4y^3-4y^4=-4y^4+10x^3+4y^3-8x^2y^3+5xy-5x\)

b) \(\left(x-3\right)\left(2x-1\right)-\left(2x-7\right)x=2x^2-7x+3-2x^2+7x=3\)

\(a,=10^3-5x+5xy-8y^3x^2+4y^3-4y^4\\ b,=3\)

tick mik nha