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

Từ giải thiết ta suy ra được: \(\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

Thay vào thì P=0

P/S: Tìm trên gg cũng có thể loại này :v

Đặt x - 2017 = a

Phương trình trên tương đương:

\(\dfrac{\left(-a\right)^2-\left(-a\right)\left(a-1\right)+\left(a-1\right)^2}{\left(-a\right)^2+\left(-a\right)\left(a-1\right)+\left(a-1\right)^2}=\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{a^2+a^2-a+a^2-2a+1}{a^2-a^2+a+a^2-2a+1}=\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3a^2-3a+1}{a^2-a+1}=\dfrac{5}{3}\)

\(\Leftrightarrow9x^2-9x+3=5x^2-5x+5\)

\(\Leftrightarrow4x^2-4x-2=0\)

\(\Leftrightarrow\left(x-\dfrac{1+\sqrt{3}}{2}\right)\left(x-\dfrac{1-\sqrt{3}}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1+\sqrt{3}}{2}\\\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)

Vậy tập nghiệm của phương trình: \(S=\left\{\dfrac{1+\sqrt{3}}{2};\dfrac{1-\sqrt{3}}{2}\right\}\)

\(2017-\left\{5^2.2^2-11\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)

Đặt : \(A=2017-\left\{5^2.2^2-11\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)

\(A=2017-\left\{25.4-11\left[49-5.8+8\left(121-121\right)\right]\right\}\)

\(A=2017-\left\{25.4-11\left[49-5.8+0\right]\right\}\)

\(A=2017-\left\{25.4-11\left[49-40\right]\right\}\)

\(A=2017-\left\{25.4-11.9\right\}\)

\(A=2017-\left\{25.4-99\right\}\)

\(A=2017-\left\{100-99\right\}\)

\(A=2017-1=2016\)

Vậy A = 2016

Bài 2:

10^n có tổng các chữ số là 1

5^3 có tổng các chữ số là 8

=>10^n+5^3 có tổng các chữ số là 9

=>10^n+5^3 chia hết cho 9

Đặt \(2x^2+x-2018=a;x^2-5x-2017=b\) ta có : 

\(a^2+4b^2=4ab\)

\(\Leftrightarrow\)\(a^2-4ab+4b^2=0\)

\(\Leftrightarrow\)\(\left(a-2b\right)^2=0\)

\(\Leftrightarrow\)\(a-2b=0\)

\(\Leftrightarrow\)\(2x^2+x-2018-2\left(x^2-5x-2017\right)=0\)

\(\Leftrightarrow\)\(2x^2+x-2018-2x^2+10x+4034=0\)

\(\Leftrightarrow\)\(11x+2016=0\)

\(\Leftrightarrow\)\(x=\frac{-2016}{11}\)

Vậy \(x=\frac{-2016}{11}\)

Chúc bạn học tốt ~