Dk: x<>-10

\(\left(4-2x\right).3=2\left(x+10\right)\)

\(12-6x=2x+20\Rightarrow12=6x+2x+20\)\(\Rightarrow12-20=8x\Rightarrow x=-\frac{8}{8}=-1\)

a) \(\frac{1}{2}-|\frac{5}{4}-2x|=\frac{1}{3}\Leftrightarrow|\frac{5}{4}-2x|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=-\frac{1}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}=\frac{13}{12}\\2x=\frac{5}{4}+\frac{1}{6}=\frac{17}{12}\end{cases}}}\)

Tự làm nốt và kết luận 

b) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)

Vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)\ne0\forall x\Rightarrow x+1=0\Leftrightarrow x=-1\)

Vậy ....

c) \(\frac{x}{y}=\frac{10}{9}\Leftrightarrow\frac{x}{10}=\frac{y}{9};\frac{y}{z}=\frac{3}{4}\Leftrightarrow\frac{y}{3}=\frac{z}{4}\Leftrightarrow\frac{y}{9}=\frac{x}{12}\)

\(\Leftrightarrow\frac{x}{10}=\frac{y}{9}=\frac{z}{12}\). Mà \(x-y+z=78\). Áp dụng t/c dãy tỉ số bằng nhau 

\(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)

\(\Rightarrow x=6.10=60;y=6.9=54;z=6.12=72\)

Vậy..........

Câu b thôi các bạn nhé, câu a mình ko cần nx với cả mình ghi sai dữ liệu câu a r

a, \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}=2x\)

\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=2x\)

\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot2\cdot3\cdot2\cdot4\cdot2\cdot5\cdot2\cdot....\cdot31\cdot2\cdot32\cdot2}=2x\)

\(\Leftrightarrow\frac{1}{2\cdot2\cdot2\cdot2\cdot....\cdot2\cdot2\cdot32}=2x\)

Có  : (31 - 1) : 1 + 1 = 31 (thừa số 2) 

\(\Rightarrow\frac{1}{2^{31}.32}=2x\)

\(\Rightarrow x=\frac{1}{2^{31}.32}\div2\)

b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Leftrightarrow x+1=x+4\)

\(\Leftrightarrow0=3\text{ (vô lý) }\)

Tìm x biết :             \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}...\frac{30}{62}.\frac{31}{64}=2x\) 

 Biến đổi mẫu thức của vế trái :     \(\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}...\frac{30}{2.31}.\frac{31}{2.32}=2x\)   Giản ước vế trái đực :

                                               \(\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2.32}=2x\)        (*)   

                                                  \(\Leftrightarrow\frac{1}{2^{36}}=2x\Leftrightarrow x=\frac{1}{2^{37}}\)  

a/ 2x - 10 - [3x - 14 - (4 - 5x) - 2x] = 2

=> 2x - 10 - (3x - 14 - 4 + 5x - 2x) = 2

=> 2x - 10 - 3x + 14 + 4 - 5x + 2x = 2

=> -4x + 6 = 0

=> -4x = -6

=> x = 3/2

b/ \(\left(\frac{1}{4}x-1\right)+\left(\frac{5}{6}x-2\right)-\left(\frac{3}{8}x+1\right)=4,5\)

\(\Rightarrow\frac{1}{4}x-1+\frac{5}{6}x-2-\frac{3}{8}x-1-\frac{9}{2}=0\)

\(\Rightarrow\frac{17}{24}x-\frac{17}{2}=0\)

\(\Rightarrow\frac{17}{24}x=\frac{17}{2}\)

\(\Rightarrow x=12\)

a, -3/4-2x= 4

 2x=-19/4

x=-19/8

(-2/3x-3/5).(-29/6)=2/5

-2/3x-3/5=-12/145

-2/3x=15/29

x=-45/58

c, x/2-3x/5+13/5=-7/5-7/10x

x/2-3x/5+7/10x=-7/5-13/5=-4

5x/10-6x/10+7x/10=-4

6x/10=-4

6x=-40

x=-20/3

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)

\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{2}{3}\)

a) x + 2x + 3x + ... +100x = -213

=>  x . (1 + 2 + 3 +... + 100) = - 213

=> x . 5050 = -213

=> x           = - 213 : 5050

=> x           = -213/5050

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

=> \(\frac{1}{2}x-\frac{1}{4}x=\frac{1}{3}-\frac{1}{6}\)

=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)

=> \(x.\frac{1}{4}=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{1}{4}\)

=> \(x=\frac{2}{3}\)

c) 3(x-2) + 2(x-1) = 10

=> 3x - 6 + 2x - 2 = 10

=> 3x + 2x - 6 - 2 = 10

=> 5x - 8 = 10

=> 5x = 10 + 8

=> 5x = 18

=> x = 18:5

=> x = 3,6

d) \(\frac{x+1}{3}=\frac{x-2}{4}\)

=> \(4\left(x+1\right)=3\left(x-2\right)\)

=>\(4x+4=3x-6\)

=> \(4x-3x=-4-6\)

=> \(x=-10\)

a)\(2x=3y,4y=5z\Leftrightarrow\frac{x}{3}=\frac{y}{2},\frac{y}{5}=\frac{z}{4}\Leftrightarrow\frac{x}{15}=\frac{y}{10},\frac{y}{10}=\frac{z}{8}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\Leftrightarrow\frac{2x}{30}=\frac{y}{10}=\frac{2z}{16}\)

ADTCDTS=NHAU TA CÓ

\(\frac{2x}{30}=\frac{y}{10}=\frac{2z}{16}=\frac{2x+y-2z}{30+10-16}=\frac{24}{24}=1\)

x=15

y=10

z=8

b) Ta có BCNN(2,3,4)=12

\(\Rightarrow\frac{2x}{12}=\frac{3x}{12}=\frac{4z}{12}\Leftrightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)

\(\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\Leftrightarrow\frac{x^2}{36}=\frac{y^2}{16}=\frac{z^2}{9}\)

ADTCDTS=NHAU TA CÓ

\(\frac{x^2}{36}=\frac{y^2}{16}=\frac{z^2}{9}=\frac{x^2+y^2+z^2}{36+16+9}=\frac{61}{61}=1\)

\(\frac{x^2}{36}=1\Rightarrow x^2=36\Rightarrow x=+_-6\)

\(\frac{y^2}{16}=1\Rightarrow x=+_-4\)

\(\frac{z^2}{9}=1\Rightarrow z=+_-3\)

TUỰ KẾT LUẬN NHA BẠN

C)\(\frac{x-6}{3}=\frac{y-8}{4}=\frac{z-10}{5}\Leftrightarrow\frac{x^2-36}{9}=\frac{y^2-64}{16}=\frac{z^2-100}{25}\)

ADTCDTS=NHAU TA CÓ

\(\frac{x^2-36}{9}=\frac{y^2-64}{16}=\frac{z^2-100}{25}=\frac{\left(x^2-36\right)+\left(y^2-64\right)+\left(z^2-100\right)}{9+16+25}\)

\(=\frac{x^2-36+y^2-64+z^2-100}{50}=\frac{\left(x^2+y^2+z^2\right)-\left(36-64-100\right)}{50}\)

\(=\frac{\left(x^2+y^2+z^2\right)-\left(36+64+100\right)}{50}=\frac{200-200}{50}=\frac{0}{50}=0\)

\(\Rightarrow\frac{x^2-36}{9}=0\Rightarrow x^2-36=0\Rightarrow x^2=36\Rightarrow x=+_-6\)

\(\frac{y^2-64}{16}=0\Rightarrow y^2-64=0\Rightarrow y^2=64\Rightarrow y==+_-8\)

\(\frac{z^2-100}{25}=0\Rightarrow z^2-100=0\Rightarrow z^2=100\Rightarrow z=+_-10\)

TỰ KẾT LUẠN NHA

\(\Rightarrow\frac{4x}{8}=\frac{2y}{3}=\frac{3z}{4}=\frac{4x-2y+3z}{8-3+4}=\frac{2\left(2x-y\right)+3z}{9}=\frac{20+3z}{9}\) (áp dụng t/c dãy tỷ số = nhau)

\(\Rightarrow\frac{3z}{4}=\frac{20+3z}{9}\Rightarrow z=\frac{16}{3}\)

\(\Rightarrow\frac{x}{2}=\frac{3}{4}.z=\frac{3}{4}.\frac{16}{3}=4\Rightarrow x=8\)

\(\Rightarrow\frac{2}{3}.y=\frac{x}{2}=\frac{8}{2}=4\Rightarrow y=6\)