A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)

\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)

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

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

b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)

\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)

\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)

 \(=\frac{1}{3.2}-\frac{5.2}{7.3}\)

\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)

\(=\frac{7}{42}-\frac{20}{42}\)

\(=-\frac{13}{42}\)

cs ng làm đung r

đag định lm

C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)

c=\(\frac{1}{1}-\frac{1}{10}\)

c=\(\frac{9}{10}\)

còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!

Em tham khảo tại đây nhé.

Câu hỏi của Phạm Minh Tuấn - Toán lớp 8 - Học toán với OnlineMath

Còn bài số 2 thì sao cô??

Ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\)

Do x + y + z = 0 nên \(\left(x+y+z\right)^2=0\)

\(\Rightarrow x^2+y^2+z^2=-2xy-2yz-2zx\)

Vậy thì \(A=\frac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)}\)

\(A=\frac{-2\left(xy+yz+zx\right)}{-4\left(xy+yz+zx\right)-2\left(xy+yz+zx\right)}\)

\(A=\frac{-2\left(xy+yz+zx\right)}{-6\left(xy+yz+zx\right)}=\frac{1}{3}\)

Cái thứ 2 là b. (a^2+c^2) đúng ko bạn

đúng rồi nha

Bạn giúp mình với

a, \(5\dfrac{4}{13}.15\dfrac{3}{41}-5\dfrac{4}{13}.2\dfrac{3}{41}\)

\(=\left(15\dfrac{3}{41}-2\dfrac{3}{41}\right).\dfrac{69}{13}=\dfrac{13.69}{13}=69\)

b, \(\dfrac{2^3}{3^3}:\dfrac{16}{27}+\dfrac{2017}{2018}-\dfrac{1}{2}.2017^0\)

\(=\dfrac{8}{27}:\dfrac{16}{27}+\dfrac{2017}{2018}-\dfrac{1}{2}.1=\dfrac{1}{2}+\dfrac{2017}{2018}-\dfrac{1}{2}=\dfrac{2017}{2018}\)

c, \(3:\left(-\dfrac{3}{2}\right)^2+\dfrac{1}{9}.\sqrt{36}=3:\dfrac{9}{4}+\dfrac{1}{9}.6=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2\)

a,

\(5\dfrac{4}{13}.14\dfrac{3}{41}-5\dfrac{4}{13}.2\dfrac{3}{41}=5\dfrac{4}{13}.\left(14\dfrac{3}{41}-2\dfrac{3}{41}\right)\)

=\(5\dfrac{4}{13}.13\)

=\(\dfrac{69}{13}.13\)

=69

     \(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+2abc=0\)

\(\Rightarrow ab^2+ac^2+bc^2+ba^2+c\left(a+b\right)^2=0\)

\(\Rightarrow ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2=0\)

\(\Rightarrow\left(a+b\right)\left(ab+c^2+ca+cb\right)=0\)

\(\Rightarrow\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

Từ đó a = -b hoặc b = -c hoặc c = -a

Nếu a = -b mà \(a^3+b^3+c^3=1\Rightarrow\left(-b\right)^3+b^3+c^3=1\Rightarrow c^3=1\Rightarrow c=1\)

Khi đó: \(A=\frac{1}{\left(-b\right)^{2017}}+\frac{1}{b^{2017}}+\frac{1}{1^{2017}}=0+1=1\)

Tương tự với các trường hợp b = -c và a = -c, ta tính được A = 1

`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`

`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`

`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`

`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`

`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`