1) 2^{2x + 1} = 32

=> 2^{2x + 1} = 2⁵

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 2

(2) 3.x³ - 100 = 275

=> 3x³ = 275 + 100

=> 3x³ = 375

=> x³ = 375 : 3

=> x³ = 125

=> x³ = 5³

=> x = 5

(4) (x - 1)³ - 2⁵ = 7²

=> (x - 1)³ = 49 + 32

=> (x - 1)³ = 81

(xem lại đề)

5) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

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

=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.5=10\end{cases}}\)

Vậy ...

6) Ta có: \(\frac{x}{2}=\frac{y}{3}\) => \(\frac{x}{10}=\frac{y}{15}\)

       \(\frac{y}{5}=\frac{z}{4}\) => \(\frac{y}{15}=\frac{z}{12}\)

=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

 \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=\frac{-49}{37}\)

=> \(\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}}\) => \(\hept{\begin{cases}x=-\frac{49}{37}\cdot10=\frac{-490}{37}\\y=-\frac{49}{37}\cdot15=-\frac{735}{37}\\z=-\frac{49}{37}\cdot12=-\frac{588}{37}\end{cases}}\)

Vậy ...

mk lm bài mà mk cho là ''khó'' nhất thôi nha 

\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và \(x+y+z=-49\)

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{15}=\frac{z}{12}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

ADTC dãy tỉ số bằng nhau ta có 

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=-\frac{49}{37}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{49}{37}.10=-\frac{490}{37}\\y=-\frac{49}{37}.15=-\frac{735}{37}\\z=-\frac{49}{37}.12=-\frac{588}{37}\end{cases}}}\)

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

<=> \(\frac{3}{2}+x=\frac{-11}{2}\)

<=> \(x=-7\)

\(\frac{1}{4}+\frac{1}{3}:3x=-5\)

<=> \(\frac{1}{3}:3x=\frac{-21}{4}\)

<=> \(3x=\frac{-4}{63}\)

<=> \(x=\frac{4}{189}\)

\(\frac{4}{5}.x=\frac{8}{35}\)

<=> \(x=\frac{2}{7}\)

\(\frac{2}{3x}-\frac{1}{4}=\frac{7}{1}\)

<=> \(\frac{2}{3x}=\frac{29}{4}\)

=> \(8=87x\)

<=> \(x=\frac{8}{87}\)

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

<=> \(\frac{3}{5x}=\frac{-5}{14}\)

<=> \(-25x=42\)

<=> \(x=\frac{-42}{25}\)

\(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16.\frac{2}{3}\right)=0\)

<=> \(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{-32}{3}=0\)

<=> \(\frac{43}{8}+x-\frac{173}{24}=\frac{-32}{3}\)

<=> \(\frac{43}{8}+x=\frac{-83}{24}\)

<=> \(x=\frac{-53}{6}\)

học tốt

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

1) x = 4/5 - 1/3

x = 7/15

2) 5/3.x=1/21

x=1/35

3) -12/13.x = 1/13

x=-1/12

7) th1: x-1=2/3

x = 5/3

Th2: x - 1 = -2/3

x=1/3

6: =>x=9/10+1/5=9/10+2/10=11/10

7: =>x=3/8-5/12=36/96-40/96=-1/24

8: =>x=7/6-5/4=14/12-15/12=-1/12

9: =>x=1/35+2/7=1/35+10/35=11/35

10: =>x=2/7-7/10=20/70-49/70=-29/70

1, \(x=-\dfrac{5}{3}-\dfrac{1}{2}=-\dfrac{10}{6}-\dfrac{3}{6}=-\dfrac{13}{6}\)

2, \(x=\dfrac{1}{3}-\dfrac{3}{5}=\dfrac{5-9}{15}=-\dfrac{4}{15}\)

4, \(x=-\dfrac{7}{9}+\dfrac{4}{3}=\dfrac{-7+12}{9}=\dfrac{5}{9}\)

5, \(x=\dfrac{5}{6}-\dfrac{7}{3}=\dfrac{5-14}{6}=-\dfrac{9}{6}=-\dfrac{3}{2}\)

6, \(x=\dfrac{9}{10}+\dfrac{1}{5}=\dfrac{9+2}{10}=\dfrac{11}{10}\)

7, \(x=\dfrac{3}{8}-\dfrac{5}{12}=\dfrac{9-10}{12}=-\dfrac{1}{12}\)

8, \(x=\dfrac{7}{6}-\dfrac{5}{4}=\dfrac{14-15}{12}=-\dfrac{1}{12}\)

9, \(x=\dfrac{1}{35}+\dfrac{2}{7}=\dfrac{1+10}{35}=\dfrac{11}{35}\)

10, \(x=-\dfrac{7}{10}+\dfrac{2}{7}=\dfrac{-49+20}{70}=\dfrac{-29}{70}\)

a) = 17/19 - 17/19 + 27/35 + 35/35 = 0 + 62/35

b) = 1/3 x 4/5 + 1/3 x6/5 + 1/3 x 2 = 1/3(4/5 + 6/5 + 2) = 1/3 x 4 = = 4/3

c) 4/7 x 2/9 + 4/7 x 7/9 + 2/3 = 4/7 x (2/9 + 7/9) + 2/3 = 4/7 x 1 + 2/3 = 26/21

A) 17/19 - 17/19 + 27/35 + 35/35 = 0 + 62/35

B) 1/3 x 4/5 + 1/3 x 6/5 + 1/3 x 2 = 1/3 x(4/5 + 6/5 x 2 ) = 1/3 x 4 = 4/3

c) TƯƠNG TỰ CÂU A VÀ B

* HOKTOT*

NHA