\(\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\left(\dfrac{1}{12}+\dfrac{2}{4}\right)\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\left(\dfrac{1}{12}+\dfrac{6}{12}\right)\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\dfrac{7}{12}\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{36}{60}\right)-\dfrac{35}{60}\right]\\ =\dfrac{3}{4}-\left(-\dfrac{71}{60}\right)\\ =\dfrac{3}{4}+\dfrac{71}{60}\\ =\dfrac{35}{60}+\dfrac{71}{60}\\ =\dfrac{106}{60}\\ =\dfrac{53}{30}\)

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\(-\dfrac{6}{7}.\dfrac{21}{12}\\ =-\dfrac{126}{84}\\ =-\dfrac{63}{42}\\ =-\dfrac{31}{14}\)

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\(\left(-5\right).\left(-\dfrac{6}{20}\right)\\ =\dfrac{\left(-5\right).\left(-6\right)}{20}\\ =\dfrac{30}{20}\\ =\dfrac{3}{2}\)

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\(-\dfrac{31}{36}:\left(-\dfrac{37}{72}\right)\\ =-\dfrac{31}{36}.\left(-\dfrac{72}{37}\right)\\ =\dfrac{2322}{1332}\)

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\(\left(-\dfrac{5}{17}\right):\left(-15\right)\\ =\left(-\dfrac{5}{17}\right).\left(-\dfrac{1}{15}\right)\\ =\dfrac{5}{255}\\ =\dfrac{1}{51}\)

\(\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\left(\dfrac{1}{12}+\dfrac{2}{4}\right)\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\left(\dfrac{1}{12}+\dfrac{6}{12}\right)\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\dfrac{7}{12}\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{36}{60}\right)-\dfrac{35}{60}\right]\\ =\dfrac{3}{4}-\left(-\dfrac{71}{60}\right)\\ =\dfrac{3}{4}+\dfrac{71}{60}\\ =\dfrac{35}{60}+\dfrac{71}{60}\\ =\dfrac{106}{60}\\ =\dfrac{53}{30}\)

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\(-\dfrac{6}{7}.\dfrac{21}{12}\\ =-\dfrac{126}{84}\\ =-\dfrac{63}{42}\\ =-\dfrac{31}{14}\)

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\(\left(-5\right).\left(-\dfrac{6}{20}\right)\\ =\dfrac{\left(-5\right).\left(-6\right)}{20}\\ =\dfrac{30}{20}\\ =\dfrac{3}{2}\)

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Hằng đẳng thức bậc cao sử dụng nhị thức newton

Theo đề ra, ta có: \(\frac{x-1}{y+2}=\frac{3}{5}\)

\(\Rightarrow\frac{x-1}{3}=\frac{y+2}{5}=\frac{x-1+y+2}{8}=\frac{23-1+2}{8}=\frac{24}{8}=3\)

\(\frac{x-1}{3}=3\Rightarrow x=3.3+1=10\)

\(\frac{y+2}{5}=3\Rightarrow y=5.3-2=13\)

Gọi x,y là nghiệm của phương trình:

\(\left\{{}\begin{matrix}S=x+y=3\\P=x.y=2\end{matrix}\right.\Rightarrow a^2-S.a+P=0\)

\(\Leftrightarrow a^2-3a+2=0\Leftrightarrow\left[{}\begin{matrix}a_1=x=2\\a_2=y=1\end{matrix}\right.\)

a)\(x^2+y^2=1^2+2^2=5\)

b)\(x^3+y^3=1^3+2^3=9\)

c)\(x^4+y^4=1^4+2^4=17\)

d)\(x^5+y^5=1^5+2^5=33\)

e)\(x^6+y^6=1^6+2^6=65\)

a: 3x=7y

=>x/7=y/3=(x-y)/(7-3)=-16/4=-4

=>x=-28; y=-12

b: x/6=y/5

=>x/6=2y/10=(x+2y)/(6+10)=20/16=5/4

=>x=30/4=15/2; y=25/4

c: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{5}=\dfrac{2x+3y+5z}{2\cdot2+3\cdot\left(-3\right)+5\cdot5}=\dfrac{6}{20}=\dfrac{3}{10}\)

=>x=3/5; y=-9/10; z=3/2

d: x/2=y/3

=>x/8=y/12

y/4=z/5

=>y/12=z/15

=>x/8=y/12=z/15

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

=>x=16; y=24; z=30

2:

a: 5/x-y/3=1/6

=>\(\dfrac{15-xy}{3x}=\dfrac{1}{6}\)

=>\(\dfrac{30-2xy}{6x}=\dfrac{x}{6x}\)

=>30-2xy=x

=>x(2y+1)=30

=>(x;2y+1) thuộc {(30;1); (-30;-1); (10;3); (-10;-3); (6;5); (-6;-5)}

=>(x,y) thuộc {(30;0); (-30;-1); (10;1); (-10;-2); (6;2); (-6;-3)}

b: x/6-2/y=1/30

=>\(\dfrac{xy-12}{6y}=\dfrac{1}{30}\)

=>\(\dfrac{5xy-60}{30y}=\dfrac{y}{30y}\)

=>5xy-60=y

=>y(5x-1)=60

=>(5x-1;y) thuộc {(-1;-60); (4;15); (-6;-10)}(Vì x,y là số nguyên)

=>(x,y) thuộc {(0;-60); (1;15); (-1;-10)}

1)\(\left(x+1\right).\left(y-2\right)=0\)                                       \(\left(x,y\inℤ\right)\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)

2)\(\left(x-5\right).\left(y-7\right)=1\)

3)\(\left(x+4\right).\left(y-2\right)=2\)

4)\(\left(x-4\right).\left(y+3\right)=-3\)

5)\(\left(x+3\right).\left(y-6\right)=-4\)

6)\(\left(x-8\right).\left(y+7\right)=5\)

7)\(\left(x+7\right).\left(y-3\right)=-6\)

8)\(\left(x-6\right).\left(y+2\right)=7\)

ok :)