ma doi

Nguyễn Khánh Phương

Bài 1 :

a) 149 - ( 35 : x + 3 ) x 17 = 13

              ( 35 : x + 3 ) x 17 = 149 - 13 

               ( 35 : x + 3 ) x 17 = 136

                ( 35 : x + 3 )       = 136 : 17

                  ( 35 : x + 3 )     = 8

                    35 - x             = 8 - 3 

                    35 - x             = 5

                            x = 35 - 5

                            x = 30

b, 121 : 11 − ( 4x + 5 )  : 3 = 4
11 − 4x + 5 : 3 = 4
4x + 5 : 3 = 11 − 4
4x + 5 : 3 = 7
4x + 5 = 7 x 3
4x + 5 = 21
4x = 21 − 5
4x = 16

x = 16 : 4

  x = 4

a,

149-(35:x+3)x17=13

(35:x+3)x17=149-13=136

35:x+3=136:17=8

35:x=8-3=5

x=35:5=7

b,

121:11-(4x+5):3=4

11-(4x+5):3=4

(4x+5):3=11-4=7

4x+5=7x3=21

4x=21-5=16

x=16:4=4

c,

mình không biết

d,\(x^{2017}=x^{2018}\)

=>x=1 hoặc x=0

\(2^x.2^{x+1}=32\)

\(\Rightarrow2^{2x+1}=2^5\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=4\)\(\Rightarrow x=2\)

Vậy \(x=2\)

Học tốt

TL:

a.\(2^6.2^n=2^{11}\) 

 \(2^{6+n}=2^{11}\) 

\(\Rightarrow n=5\) 

b. \(3^7:3^n=3^4\) 

  \(3^{7-n}=3^4\) 

\(\Rightarrow n=3\) 

c.\(2^n.32=2^{10}\)

 \(2^{n+5}=2^{10}\) 

\(\Rightarrow n=5\) 

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

=> \(\frac{1}{7}x=\frac{3}{7}-\frac{2}{3}=-\frac{5}{21}\)

=> \(x=-\frac{5}{21}:\frac{1}{7}=-\frac{5}{21}\cdot7=-\frac{5}{3}\)

b) \(3x^2-2=72\)=> 3x² = 74 => x² = 74/3 => x không thỏa mãn

c) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)

=> \(\left(19x+2\cdot25\right):14=5^2-4^2=9\)

=> \(\left(19x+50\right):14=9\)

=> \(19x+50=126\)

=> \(19x=76\)

=> x = 4

d) \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{16}+x:\frac{1}{32}=343\)

=> \(x\cdot2+x\cdot4+x\cdot8+x\cdot16+x\cdot32=343\)

=> \(x\left(2+4+8+16+32\right)=343\)

=> x . 62 = 343

=> x = 343/62

a)Đặt k, ta có:

x/2=k =>2k=x; y/3=k =>3k=y; z/5=k =>5k=z

thay x/2=k =>2k=x; y/3=k =>3k=y; z/5=k =>5k=z vào x²+y²+z²=152, tao có:

(2k)²+(3k)²+(5k)²=152

=>4xk²+9xk²+25xk²=152

=>k²x38=152

=>k²=4=>k=2 hoặc k=-2

Với k=2

=>x=4;y=6;z=10

Với k=-2

=>x=-4;y=-6;z=-10

Vậy (x=4;y=6;z=10) hoặc (x=-4;y=-6;z=-10)

b)Áp dụng dãy tỉ số bằng nhau, ta có :

x/4=y/7=z/9=(2x)/8=(2x-y)/8-7=2

=>x=8;y=14;z=18

Vậy........

=>x-25=40

hay x=65

x-25=40

hay x=65

b \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

hay \(x\in\left\{0;2\right\}\)

c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)

=>(x-8)(3x+2)=0

=>x=8 hoặc x=-2/3

d: \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>x=2 hoặc x=1

e: \(\Leftrightarrow x\left(x^2-11x+30\right)=0\)

=>x(x-5)(x-6)=0

hay \(x\in\left\{0;5;6\right\}\)

b: \(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

hay \(x\in\left\{0;2\right\}\)

c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)

=>(x-8)(3x+2)=0

hay \(x\in\left\{8;-\dfrac{2}{3}\right\}\)

d: \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

=>x=1 hoặc x=2