Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=kb;c=c=kd\)

Ta có: \(\dfrac{a^2+ac}{c^2-ac}=\dfrac{b^2k^2+kb.kd}{d^2k^2-kb.kd}=\dfrac{bk^2\left(b+d\right)}{dk^2\left(d-b\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(1\right)\)

\(\dfrac{b^2+bd}{d^2-bd}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\dfrac{a^2+ac}{c^2-ac}=\dfrac{b^2+bd}{d^2-bd}\) 

Bài 1

a) \(A=\dfrac{1}{2}.\dfrac{1}{-3}+\dfrac{1}{-3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{-5}+\dfrac{1}{-5}.\dfrac{1}{6}\)

\(=-\left(\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{5}+\dfrac{1}{5}.\dfrac{1}{6}\right)\)

\(=-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)\)

\(=-\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)

\(=-\dfrac{1}{3}\)

b) \(B=3\sqrt{\left(-5\right)^2}-0,5.\dfrac{1}{3}.\sqrt{9}+\left|-\dfrac{2^2}{3}\right|:\left(-1\dfrac{1}{3}\right)\)

\(=3.5-0,5.\dfrac{1}{3}.3+\dfrac{4}{3}:\left(-\dfrac{4}{3}\right)\)

\(=15-0,5-1\)

\(=13,5\)

c) \(C=\dfrac{\left(-2\right)^5.6^4+9^2.8^4}{\left(-12\right)^4}\)

\(=\dfrac{\left(-2\right)^5.2^4.3^4+\left(3^2\right)^2.\left(2^3\right)^4}{12^4}\)

\(=\dfrac{-2^9.3^4+3^4.2^{12}}{\left(2^2.3\right)^4}\)

\(=\dfrac{2^9.3^4.\left(-1+2^3\right)}{2^8.3^4}\)

\(=2.7\)

\(=14\)

Bài 2

a) \(\dfrac{1}{3}-\dfrac{1}{3}:\left|2x-1\right|=-\dfrac{2}{3}\)

\(\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}-\left(-\dfrac{2}{3}\right)\)

\(\dfrac{1}{3}:\left|2x-1\right|=1\)

\(\left|2x-1\right|=\dfrac{1}{3}:1\)

\(\left|2x-1\right|=\dfrac{1}{3}\)

\(\Rightarrow2x-1=\dfrac{1}{3};2x-1=-\dfrac{1}{3}\)

*) \(2x-1=\dfrac{1}{3}\)

\(2x=\dfrac{1}{3}+1\)

\(2x=\dfrac{4}{3}\)

\(x=\dfrac{4}{3}:2\)

\(x=\dfrac{2}{3}\)

*) \(2x-1=-\dfrac{1}{3}\)

\(2x=-\dfrac{1}{3}+1\)

\(2x=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}:2\)

\(x=\dfrac{1}{3}\)

Vậy \(x=\dfrac{1}{3};x=\dfrac{2}{3}\)

b) \(\dfrac{1-x}{-9}=\dfrac{-25}{1-x}\)

Điều kiện: \(x\ne1\)

\(\left(1-x\right)^2=\left(-25\right).\left(-9\right)\)

\(\left(1-x\right)^2=225\)

\(1-x=15\) hoặc \(1-x=-15\)

*) \(1-x=15\)

\(x=1-15\)

\(x=-14\) (nhận)

*) \(1-x=-15\)

\(x=1-\left(-15\right)\)

\(x=-16\) (nhận)

Vậy \(x=-16;x=-14\)

Bài 5

\(A=\dfrac{1}{7^2}-\dfrac{1}{7^4}+\dfrac{1}{7^6}-\dfrac{1}{7^8}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\)

\(\Rightarrow\dfrac{1}{7^2}.A=\dfrac{1}{7^4}-\dfrac{1}{7^6}+\dfrac{1}{7^8}-\dfrac{1}{7^{10}}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\)

\(\Rightarrow A+\dfrac{1}{49}.A=\left(\dfrac{1}{7^2}-\dfrac{1}{7^4}+\dfrac{1}{7^6}-\dfrac{1}{7^8}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\right)+\left(\dfrac{1}{7^4}-\dfrac{1}{7^6}+\dfrac{1}{7^8}-\dfrac{1}{7^{10}}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\right)\)

\(\dfrac{50A}{49}=\dfrac{1}{7^2}-\dfrac{1}{7^{102}}\)

\(A=\left(\dfrac{1}{7^2}-\dfrac{1}{7^{100}}\right):\dfrac{50}{49}\)

\(=\left(\dfrac{1}{49}-\dfrac{1}{7^{100}}\right).\dfrac{49}{50}\)

\(=\dfrac{1}{50}-\dfrac{1}{7^{98}.50}< \dfrac{1}{50}\)

Vậy \(A< \dfrac{1}{50}\)

\(\dfrac{1}{3}-\dfrac{1}{3}:\left|2x-1\right|=-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}-\left(-\dfrac{2}{3}\right)\)

\(\Rightarrow\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}+\dfrac{2}{3}\)

\(\Rightarrow\dfrac{1}{3}:\left|2x-1\right|=1\)

\(\Rightarrow\left|2x-1\right|=\dfrac{1}{3}:1\)

\(\Rightarrow\left|2x-1\right|=\dfrac{1}{3}\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=\dfrac{1}{3}\\2x-1=-\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{4}{3}\\2x=\dfrac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}:2\\x=\dfrac{2}{3}:2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{3}\end{matrix}\right.\)

a: Xet ΔMAC và ΔMEB co

MA=ME

góc AMC=góc EMB

MC=MB

=>ΔMAC=ΔMEB

b: ΔMAC=ΔMEB

=>góc MAC=góc MEB và AC=EB

=>AC//EB

c: Xét tứ giác ABEC có

AC//EB

AC=EB

=>ABEC là hình bình hành

mà AB=BE

nên ABEC là hình thoi

=>AM là phân giác của góc BAC

d: Xét ΔMNB vuông tại N và ΔMPC vuông tại P có

MB=MC
góc MBN=góc MCP

=>ΔMNB=ΔMPC

=>MN=MP và góc NMB=góc PMC

=>góc NMB+góc BMP=180 độ

=>N,M,P thẳng hàng

mà MN=MP

nên M là trung điểm của NP

a: Số ko chia hết cho 3 có dạng là 3k+1 hoặc 3k+2(k\(\in Z\))

b: Ư(6)={1;-1;2;-2;3;-3;6;-6}

Có `x` là `y` là `2` đại lượng TLN

`=>x=a/y`

Thay `x=5;y=10` vào `x=a/y` , ta đc :

`5=a/10`

`a=5*10`

`a=50`

`=>D`

a: Xét ΔAMB và ΔAMC co

AM chung

MB=MC

AB=AC

=>ΔAMB=ΔAMC

b: ΔABC cân tại A

mà AM là trung tuyến

nên AM vuông góc BC

c: góc FBC+góc C=90 độ

góc MAC+góc C=90 độ

=>góc FBC=góc MAC

Cảm ơn bạn Thịnh nhiều nhé. Còn câu d mình ráng giải vậy, khó quá.

y1=6y2

y1-y2=15

=>y1=18; y2=3

x và y tỉ lệ nghịch

nên x1y1=x2y2

=>x1*18=x2*3

=>x1*6=x2*1

=>x2=6x1

\(A=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\\ =\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\\ =\dfrac{1}{2}\cdot\dfrac{3}{4}:1+\dfrac{5}{8}=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=1\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=1/2*3/4+5/8

=3/8+5/8=1