a)hình như đề sai thì phải

sửa lại

\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)

=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)

=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)

Đề sai bạn nhé. Đưa dữ kiện 3 ẩn bắt tính biểu thức chứa 2 ẩn làm sao làm được ?

Bạn kiểm tra lại nha

Quốc Anh Đinh, bạn tham khảo link này nhé!

https://hoc24.vn/hoi-dap/question/217301.html

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\) \(\begin{cases} a = bk \\ c = dk \end{cases}\)

Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(1\right)\)

\(\dfrac{a.c}{b.d}=\dfrac{bk.dk}{b.d}=\dfrac{k^2.b.d}{b.d}=k^2\left(2\right)\)

Từ (1) và (2) suy ra: \(\dfrac{a.c}{b.d}=\dfrac{a^2+c^2}{b^2+d^2}\) \(\rightarrow đpcm\).

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

\(\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)

\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)

=>-6a+5b=6a-5b

=>-12a=-10b

=>6a=5b

hay a/b=5/6

\(VT=\dfrac{a+c}{a+b}+\dfrac{b+d}{b+c}+\dfrac{c+a}{c+d}+\dfrac{d+b}{d+a}\)

\(=\left(a+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{c+d}\right)+\left(b+d\right)\left(\dfrac{1}{b+c}+\dfrac{1}{d+a}\right)\)

Ap dụng \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y} \left(\forall x,y>0\right)\)

Ta có: \(VT\ge\left(a+c\right).\dfrac{4}{a+b+c+d}+\left(b+d\right).\dfrac{4}{a+b+c+d}\)

\(=\dfrac{4\left(a+b+c+d\right)}{\left(a+b+c+d\right)}=4\left(ĐPCM\right)\)

Từ a/b=c/d⇒a/c=b/d

Áp dụng tính chất dãy tỉ số bằng nhau

a/c=b/d=a+b/c+d

⇒a^3/c^3=b^3/d^3=(a+b)^3/(c+d)^3 (1)

Từ a^3/c^3=b^3/d^3=a^3-b^3/c^3-d^3 (2)

Từ (1) và (2)

⇒(a+b)^3/(c+d)^3=a^3-b^3/c^3-d^3

Từ \(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)

\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\\\dfrac{1}{c}+\dfrac{1}{a}=\dfrac{1}{a}+\dfrac{1}{b}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\\\dfrac{1}{c}=\dfrac{1}{b}\end{matrix}\right.\)\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\Rightarrow a=b=c\)

Khi đó \(P=\dfrac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\dfrac{3a^3}{3a^3}=1\)

c, \(\left(7-3x\right)\left(2x+1\right)=0\)

=> \(7-3x=0\) hoặc \(2x+1=0\)

\(3x=7-0\) hoặc \(2x=0-1\)

\(3x=7\) hoặc \(2x=-1\)

\(x=7:3\) hoặc \(x=-1:2\)

\(x=\dfrac{7}{3}\) hoặc \(x=-0,5\)

Vậy, \(x\in\left\{\dfrac{7}{3};-0,5\right\}\)

Đăng từng bài một thôi bạn!

1)\(\left(-\dfrac{5}{13}\right)^{2017}.\left(\dfrac{13}{5}\right)^{2016}\)

\(=\left(-\dfrac{5}{13}\right).\left(-\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)

\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)

\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}.\dfrac{13}{5}\right)^{2016}\)

\(=\left(-\dfrac{5}{13}\right).1^{2016}\)

\(=-\dfrac{5}{13}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{y+x+t}=\dfrac{y}{z+t+x}=\dfrac{y}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+z=3t\\y+z+t=3x\\z+t+x=3y\\t+x+y=3z\end{matrix}\right.\)

\(\Rightarrow x=y=z=t\)

Thay vào P ta được :

\(P=1+1+1+1=4\)