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

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

<=> \(-\frac{1}{3}x-\frac{17}{12}=1\)

<=> \(-\frac{1}{3}x=\frac{29}{12}\)

<=> \(x=-\frac{29}{4}\)

\(\frac{5}{6}\left(x+2\right)-x-\frac{1}{2}=\frac{1}{3}\)

<=> \(\frac{5}{6}x+\frac{5}{3}-x-\frac{1}{2}=\frac{1}{3}\)

<=> \(-\frac{1}{6}x+\frac{7}{6}=\frac{1}{3}\)

<=> \(-\frac{1}{6}x=-\frac{5}{6}\)

<=> \(x=5\)

học tốt

–6 < x < 4

Các giá trị của \(x\) là :

-5;-4;-3;-2;-1;0;1;2;3

–3 ≤ x < 4 

Các giá trị của \(x\)là:

-3 ; -2 ; -1 ; 0 ; 1 ; 2;3

-6<x<4

\(\Rightarrow x\in\left\{-5;-4;-3;-2;-1;0;1;2;3\right\}\)

-3\(\le\)x<4

\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2;3\right\}\)

6/20:3=6/20x1/3=6/60=1/10

11h-8h30p=2h30p=2.5h

10 7/10-4 3/10

=107/10-43/10

=64/10=32/5

chúc bn học tốt!

cảm ơn cậu ạ

a)

(x+4)(3x-5) = 0

=> x + 4 = 0 hoặc 3x-5 = 0

     x = -4                 x = 5/3

b)

  2x² + 7x + 3 = 0

  2x² + 6x + x + 3= 0

  (2x+1)(x+3) = 0

=> 2x+1 = 0 hoặc x + 3 = 0

    x = -1/2              x = -3

\(4^{x+3}+4^{x+2}+4^{x+1}+4^x=5440\)

\(\Rightarrow4^x.4^3+4^x.4^2+4^x.4+4^x=5440\)

\(\Rightarrow4^x\left(4^3+4^2+4+1\right)=5440\)

\(\Rightarrow4^x.\left(64+16+4+1\right)=5440\)

\(\Rightarrow4^x.85=5440\)

\(\Rightarrow4^x=5440:85\)

\(\Rightarrow4^x=64=4^3\)

\(\Rightarrow x=3\)

dễ quá bạn ơi giải câu này nè mới chất

Q= 1² + 2² + 3² +...+ 100²

a)Tách 3=1+1+1 rồi phân phát chúng cho 3 phân số được là

\(\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{3}=0\)<=>(x+10)\(\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{3}\right)=0\)

Dễ rồi

bNguyễn Thị Mai Huyền xem có câu hỏi giống đấy

Bài 4:

a) Ta có: \(a^4+a^2+1\)

\(=a^4+2a^2+1-a^2\)

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

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

b) Ta có: \(a^4+a^2-2\)

\(=a^4+2a^2-a^2-2\)

\(=a^2\left(a^2+2\right)-\left(a^2+2\right)\)

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

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

c) Ta có: \(x^4+4x^2-5\)

\(=x^4+5x^2-x^2-5\)

\(=x^2\left(x^2+5\right)-\left(x^2+5\right)\)

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

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

d) Ta có: \(x^3-19x-30\)

\(=x^3-25x+6x-30\)

\(=x\left(x^2-25\right)+6\left(x-5\right)\)

\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2+5x\right)+6\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2+5x+6\right)\)

\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)

\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)

\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)

e) Ta có: \(x^3-7x-6\)

\(=x^3-4x-3x-6\)

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

\(=x\left(x-2\right)\left(x+2\right)-3\left(x+2\right)\)

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

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

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

\(=\left(x+2\right)\left[x\left(x-3\right)+\left(x-3\right)\right]\)

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

f) Ta có: \(x^3-5x^2-14x\)

\(=x\left(x^2-5x-14\right)\)

\(=x\left(x^2-7x+2x-14\right)\)

\(=x\left[x\left(x-7\right)+2\left(x-7\right)\right]\)

\(=x\left(x-7\right)\left(x+2\right)\)

\(a.\left[\left(6.x-39\right).7\right].4=12\)

\(\left(6.x-39\right).7=12:4\)

\(6.x-39=3:7\)

\(6.x=\frac{3}{7}+39\)

\(x=\frac{276}{7}:6\)

\(x=\frac{46}{7}\)

( câu a )

\(b.200-\left(2.x+6\right)=4^3\)

\(200-\left(2.x+6\right)=64\)

\(2.x+6=200-64\)

\(2.x=136-6\)

\(x=130:2\)

\(x=65\)

( câu b )