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26 tháng 5 2017

Đặt x + 29 = a (a \(\ne-29;-30\))

Đề trở thành: \(\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{5}{4}\)

\(\Leftrightarrow\frac{\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}=\frac{5}{4}\)

\(\Leftrightarrow\frac{a^2+2a+1+a^2}{a^2\left(a^2+2a+1\right)}=\frac{5}{4}\)

\(\Leftrightarrow\frac{2a^2+2a+1}{a^4+2a^3+a^2}=\frac{5}{4}\)

\(\Leftrightarrow8a^2+8a+4=5a^4+10a^3+5a^2\)

\(\Leftrightarrow5a^4+10a^3-3a^2-8a-4=0\)

\(\Leftrightarrow5a^4+10a^3-3a^2-6a-2a-4=0\)

\(\Leftrightarrow5a^3\left(a+2\right)-3a\left(a+2\right)-2\left(a+2\right)=0\)

\(\Leftrightarrow\left(a+2\right)\left(5a^3-3a-2\right)=0\)

\(\Leftrightarrow\left(a+2\right)\left(5a^3-5a+2a-2\right)=0\)

\(\Leftrightarrow\left(a+2\right)\left(a-1\right)\left(5a^2+5a+2\right)=0\)

tới đây dễ r`

26 tháng 5 2017

x = -28, đoán bừa.

4 tháng 6 2016

Tìm x, biết:

3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x2;5;10;17)

2(x1)(x3) +5(x3)(x8) +12(x8)(x20) 1x20 =34 (x1;3;8;20)

x+110 +2+111 x+112 =x+113 +x+114 

x1030 +x1443 +x595 +x1488 =0

4 tháng 6 2016

Trả lời luôn à bạn

16 tháng 8 2019

1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

16 tháng 8 2019

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

8 tháng 1 2020

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

14 tháng 10 2020

a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)

     \(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) =>   \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\)  =>   \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) =>   \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)

Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)

b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)\(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)

=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=>  \(\frac{27x}{4}=\frac{27}{40}\)

\(27x.40=27.4\)

\(1080.x=108\)

             \(x=\frac{1}{10}\)

Vậy \(x=\frac{1}{10}\)

c) \(\left|x-1\right|+4=6\)

\(\left|x-1\right|=6-4\)

\(\left|x-1\right|=2\)

\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=>  \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Vậy \(x=\left[3,-1\right]\)

d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)

e) \(\left(x^2-3\right)^2=16\)

\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)

\(x^2=7=>x=\sqrt{7}\)

Vậy \(x=\sqrt{7}\)

f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)

               \(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\) 

               \(\frac{2}{5}x=-\frac{4}{15}\)

          \(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)

Vậy \(x=-\frac{2}{3}\)

g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)

\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)

\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)

Vậy \(x=-3\)

k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)

\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)

\(\frac{2}{5}x=\frac{4}{15}\)

      \(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)

Vậy \(x=\frac{2}{15}\)

I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)

\(\frac{3}{5}x=\frac{5}{14}\)

\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)

Vậy \(x=\frac{25}{42}\)

14 tháng 12 2016

B1:Ta có:

2800=(28)100=256100(1)

3600=(36)100=729100(2)

Từ (1),(2)

Ta có: 256100<729100(Vì 256<729)

=>2800<3600

15 tháng 12 2016

Mình làm bài 1 thôi nhé!

Ta có :2^800=2^2.400=(2^2)^400=4^400=4^4.100=(4^4)^100=256^100

           3^600=3^2.300=(3^2)^300=9^300=9^3.100=(9^3)^100=729^100

Vì 256^100<729^100 nên 2^800<3^600

^ là mũ nhé!