\(\frac{1}{3}\le\frac{a^2-2a+4}{a^2+2a+...">
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28 tháng 11 2016

Bài 2:

a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)

Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)

\(\Rightarrow6x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)

\(\Rightarrow4x+12=6x\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\)

Vậy x = 6

b) Giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)

\(=\frac{14-5}{8}=\frac{9}{8}\)

+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)

+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)

+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)

Vậy ...

c) \(5^x+5^{x+1}+5^{x+2}=3875\)

\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)

\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)

\(\Rightarrow5^x.31=3875\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

Vậy x = 3

28 tháng 11 2016

@@ good :D

17 tháng 10 2020

1) Ta có: \(\frac{a}{b}=\frac{c}{d}\)

\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Leftrightarrow\frac{a}{c}+1=\frac{b}{d}+1\)

\(\Leftrightarrow\frac{a+c}{c}=\frac{b+d}{d}\)(đpcm)

2) Để \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\) thì \(\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)

\(\Leftrightarrow\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a}{2c}=\frac{3b}{3d}\)

\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{a}{c}=\frac{b}{d}\)

\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

hay \(\frac{a}{b}=\frac{c}{d}\)(đpcm)

3) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\frac{ab}{cd}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\)(1)

Ta có: \(\frac{a^2-b^2}{c^2-d^2}\)

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

Từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

4) Ta có: \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

nên \(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2\cdot k^2+b^2}{d^2\cdot k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(3)

Ta có: \(\left(\frac{a+b}{c+d}\right)^2\)

\(=\left(\frac{bk+b}{dk+d}\right)^2\)

\(=\left(\frac{b\left(k+1\right)}{d\left(k+1\right)}\right)^2\)

\(=\left(\frac{b}{d}\right)^2=\frac{b^2}{d^2}\)(4)

Từ (3) và (4) suy ra \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)

29 tháng 8 2015

quá dễ              

29 tháng 8 2015

dễ ngon lm đi                      

11 tháng 3 2020

Xet \(M-1=a+\frac{2a+2}{2-b}-\left(\frac{2a-b}{2+b}+1\right)+\frac{4a}{b^2-4}\)

=\(a+\left(2a+2\right)\left(\frac{1}{2-b}-\frac{1}{2+b}\right)+\frac{4a}{b^2-4}\)

=\(\frac{ab^2-4a-4ab-4b+4a}{b^2-4}\)

=\(\frac{ab^2-4ab-4b}{b^2-4}\)

den doan nay em xet rieng tu so \(ab^2-4ab-4b\)

thay b=a/a+1 vao \(\frac{a^3}{\left(a+1\right)^2}-\frac{4a^2}{a+1}-\frac{4}{a+1}\)

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

xet mau so b^2-4=(a/a+1)^-4

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

den day thay vao la xong nha

8 tháng 4 2020

i don't know ok

26 tháng 1 2017

Bài 2:

Giải:

Đặt \(\frac{x}{5}=\frac{y}{4}=k\Rightarrow x=5k,y=4k\)

Ta có: \(x^2-y^2=1\)

\(\Rightarrow\left(5k\right)^2-\left(4k\right)^2=1\)

\(\Rightarrow5^2.k^2-4^2.k^2=1\)

\(\Rightarrow k^2\left(5^2-4^2\right)=1\)

\(\Rightarrow k^2.9=1\)

\(\Rightarrow k^2=\frac{1}{9}\)

\(\Rightarrow k=\pm\frac{1}{3}\)

+) \(k=\frac{1}{3}\Rightarrow x=\frac{5}{3};y=\frac{4}{3}\)

+) \(k=\frac{-1}{3}\Rightarrow x=\frac{-5}{3};y=\frac{-4}{3}\)

Vậy cặp số \(\left(x;y\right)\)\(\left(\frac{5}{3};\frac{4}{3}\right);\left(\frac{-5}{3};\frac{-4}{3}\right)\)

Bài 3:

Giải:

Ta có: \(2a=3b\Rightarrow\frac{a}{3}=\frac{b}{2}\Rightarrow\frac{a}{21}=\frac{b}{14}\)

\(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{b}{21}=\frac{c}{15}\)

\(\Rightarrow\frac{a}{21}=\frac{b}{14}=\frac{c}{15}\)

...

Bài 4:

Giải:

Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)

\(\Rightarrow a=2k,b=3k,c=5k\)

Ta có: \(P=\frac{b+c-a}{a-b+c}=\frac{3k+5k-2k}{2k-3k+5k}=\frac{\left(3+5-2\right)k}{\left(2-3+5\right)k}=\frac{6}{4}=\frac{3}{2}\)

Vậy \(P=\frac{3}{2}\)

26 tháng 1 2017

4) đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\)

=> a = 2k

b = 3k

c = 4k

thay vào P ta có:

P = \(\frac{3k+4k-2k}{2k-3k+4k}=\frac{7k-2k}{4k-k}=\frac{5k}{3k}=\frac{5}{3}\)

vậy P = \(\frac{5}{3}\)

27 tháng 9 2017

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

a)\(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(1)

\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(2)

Từ (1) và (2) \(\Rightarrow\)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

b)\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(1)

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

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

c)\(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(bk+b\right)^2}{\left(ck+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2}{d^2}\)(1)

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

Từ (1) và(2)\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)

k cho mình nhé