a) Đặt :\(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\)

\(\Rightarrow a=kb;c=kd\)

Thay: \(\frac{a^2-b^2}{c^2-d^2}=\frac{k^2.b^2-b^2}{k^2.d^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\) ⁽¹⁾

Mặt khác: \(\frac{ab}{cd}=\frac{kb^2}{kd^2}=\frac{b^2}{d^2}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\left(đpcm\right)\) 

b) \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{b^2.\left(k-1\right)^2}{d^2.\left(k-1\right)^2}=\frac{b^2}{d^2}\) ⁽³⁾

Từ (2) và (3) \(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\left(đpcm\right)\)

a.

Đặt \(k=\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow a=bk;c=dk\)

\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{k^2b^2-b^2}{k^2d^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}=\frac{b}{d}\cdot\frac{b}{d}\)

Mà \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Thay vào ta được \(đpcm\)

b tương tự 

1

\(A=\frac{2019^{2019}+1}{2019^{2020}+1}< \frac{2019^{2019}+1+2018}{2019^{2020}+1+2018}=\frac{2019^{2019}+2019}{2019^{2020}+2019}=\frac{2019\left(2019^{2018}+1\right)}{2019\left(2019^{2019}+1\right)}\)

\(=\frac{2019^{2018}+1}{2019^{2019}+1}\)

2

\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)

\(=\frac{100^{100}+1}{100^{99}+1}=N\)

ta có:\(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\frac{b}{a}=\frac{c}{d}\)

                               \(\Rightarrow1-\frac{b}{a}=1-\frac{c}{d}\)

                                \(\Rightarrow\frac{a}{a}-\frac{b}{a}=\frac{c}{c}-\frac{d}{c}\)

                                \(\Rightarrow\frac{a-b}{a}=\frac{c-d}{c}\)

                            hay: \(\frac{a}{a-b}=\frac{c}{c-d}\)(đpcm)

Cách 1 : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{a-b}{c-d}\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)

Cách 2 : \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow ac-ad=ac-bc\)

\(\Rightarrow a(c-d)=c(a-b)\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)

Cách 3 : Đặt \(\frac{a}{b}=\frac{c}{d}=m\Rightarrow a=mb,c=md\)

Ta có : \(\frac{a}{a-b}=\frac{mb}{mb-b}=\frac{mb}{b(m-1)}=\frac{m}{m-1}\)

\(\frac{c}{c-d}=\frac{md}{md-d}=\frac{md}{d(m-1)}=\frac{m}{m-1}\)

Do đó : \(\frac{a}{a-b}=\frac{c}{c-d}\)

Cách 4 : \(\frac{a}{a-b}=\frac{c}{c-d}\Rightarrow a(c-d)=c(a-b)\)

\(\Rightarrow ac-ad=ac-bc\Rightarrow ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\) đẳng thức đúng

Do đó , ta có : \(\frac{a}{a-b}=\frac{c}{c-d}\)là đẳng thức đúng.

a) (5x+1)² - (5x+3)(5x-3)=30

=> 25x² +50x +1 - (25x²-9)=30

=> 25x² + 50x +1 - 25x² + 9 = 30

=> 50x = 30 - 9 -1 

=> 50x = 20

=> x= 2/5

#)Giải :

a) \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)

\(\Rightarrow25x^2+10x+1-25x^2+9=30\)

\(\Rightarrow\left(25x^2-25x^2\right)+10x+1+9=30\)

\(\Rightarrow10x+10=30\)

\(\Rightarrow x=2\)

b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Rightarrow x^3-27-x\left(x^2-4\right)=15\)

\(\Rightarrow x^3-27x-x^3+4x=15\)

\(\Rightarrow4x-27=15\)

\(\Rightarrow4x=42\)

\(\Rightarrow x=\frac{21}{2}\)

Mình chỉ làm bài 1a, và bài 3 thôi nhé,còn lại là bạn tự làm nhé

Bài 1:

a, Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)

\(\Rightarrow\left[\frac{a}{b}\right]^2=\left[\frac{c}{d}\right]^2=\left[\frac{a+c}{b+d}\right]^2\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{(a+c)^2}{(b+d)^2}\Rightarrow\frac{a^2+c^2}{b^2+d^2}=\frac{(a+c)^2}{(b+d)^2}\)

Bài 3 : Sửa đề : Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)

CM : a = b = c

Cách 1 : Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)

vì \(a+b+c\ne0\)

\(\frac{a}{b}=1\Rightarrow a=b;\frac{b}{c}=1\Rightarrow b=c\)

Do đó : \(a=b=c\).

Cách 2 : Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=m\), ta có : \(a=bm,b=cm,c=am\)

Do đó : \(a=bm=m(mc)=m\left[m(ma)\right]\)

\(\Rightarrow a=m^3a\Rightarrow m^3=1(a\ne0)\Rightarrow m=1\)

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)

Cách 3 : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}=\left[\frac{a}{b}\right]^3\Rightarrow1=\left[\frac{a}{b}\right]^3\Rightarrow\frac{a}{b}=1\)

Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)

\(6.\left(\frac{2}{3}\right)^2-3.\left(-\frac{2}{3}\right)^2+1\)

\(=6.\frac{4}{9}-3.\frac{4}{9}+1\)

\(=\frac{4}{9}\left(6-3\right)+1\)

\(=\frac{4}{9}.3+1=\frac{4}{3}+1=\frac{7}{3}\)

a) \(A\left(-2;\frac{4}{5}\right)\)thuộc đồ thị hàm số nên ta có:

\(\frac{4}{5}=\left(2m+\frac{3}{5}\right)\left(-2\right)\)

<=> \(2m+\frac{3}{5}=-\frac{2}{5}\)

<=> \(2m=-1\)

\(\Leftrightarrow m=-\frac{1}{2}\)

b) Với \(m=-\frac{1}{2}\)

\(y=\left(2.\left(-\frac{1}{2}\right)+\frac{3}{5}\right)x\)

\(y=-\frac{2}{5}x\)

c) 

Ta có bảng:

Đồ thị:

d) 

Với y=4 => \(4=-\frac{2}{5}.x\Leftrightarrow x=4:\left(-\frac{2}{5}\right)=-10\)

=> \(A\left(-10;4\right)\)

Với \(y=-\frac{3}{5}\Rightarrow-\frac{3}{5}=-\frac{2}{5}x\Rightarrow x=\left(-\frac{3}{5}\right):\left(-\frac{2}{5}\right)=\frac{3}{2}\)

=> \(B\left(\frac{3}{2};-\frac{3}{5}\right)\)

e) Với \(x=-5\Rightarrow y=-\frac{2}{5}x=-\frac{2}{5}.\left(-5\right)=2\)

=> \(C\left(-5;2\right)\)

Với \(x=\frac{5}{2}\Rightarrow y=-\frac{2}{5}.x=-\frac{2}{5}.\frac{5}{2}=-1\)

=> \(D\left(\frac{5}{2};-1\right)\)

cám ơn Nguyễn Linh Chi ^^

\(x+\left[x-\frac{3}{2}\right]=0\)

\(\Leftrightarrow x+x-\frac{3}{2}=0\)

\(\Leftrightarrow2x=\frac{3}{2}\Leftrightarrow x=\frac{3}{2}:2=\frac{3}{4}\)

x + (x - \(\frac{3}{2}\)) = 0

<=> x + x - \(\frac{3}{2}\)= 0

<=> 2x                 = \(\frac{3}{2}\)

#hatsunemiku

<=> x                   = \(\frac{3}{4}\)

\(x+9\left(x-\frac{3}{2}\right)=0\)

\(\Rightarrow x+9x-\frac{27}{2}=0\)

\(\Rightarrow10x=\frac{27}{2}\)

\(\Rightarrow x=\frac{27}{2}\div10=\frac{27}{2}.\frac{1}{10}=\frac{27}{20}\)

\(x+9\left[x-\frac{3}{2}\right]=0\)

\(\Leftrightarrow x+9x-\frac{27}{2}=0\)

\(\Leftrightarrow10x=\frac{27}{2}\)

\(\Leftrightarrow x=\frac{27}{2}:10=\frac{27}{2}\cdot\frac{1}{10}=\frac{27}{20}\)

Vậy : ...