a: \(\Leftrightarrow\left(4x+12\right)\left(3x-2\right)-\left(3x+3\right)\left(4x-1\right)=-27\)

\(\Leftrightarrow12x^2-8x+36x-24-\left(12x^2-3x+12x-3\right)=-27\)

\(\Leftrightarrow12x^2+28x-24-12x^2-9x+3=-27\)

\(\Leftrightarrow19x-21=-27\)

=>19x=-6

hay x=-6/19

b: \(\left(x+1\right)\left(3x^2-x+1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow3x^3-x^2+x+3x^2-x+1+4x^2-3x^3=\dfrac{5}{2}\)

\(\Leftrightarrow6x^2+1=\dfrac{5}{2}\)

\(\Leftrightarrow6x^2=\dfrac{3}{2}\)

\(\Leftrightarrow x^2=\dfrac{3}{12}=\dfrac{1}{4}\)

=>x=1/2 hoặc x=-1/2

c: \(\Leftrightarrow2\left(x^2-4\right)-4\left(x^2-x-2\right)+\left(5x+8\right)\left(x+2\right)=0\)

\(\Leftrightarrow2x^2-8-4x^2+4x+8+5x^2+10x+8x+16=0\)

\(\Leftrightarrow3x^2+22x+16=0\)

\(\text{Δ}=22^2-4\cdot3\cdot16=292>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-22-2\sqrt{73}}{6}=\dfrac{-11-\sqrt{73}}{3}\\x_2=\dfrac{-11+\sqrt{73}}{3}\end{matrix}\right.\)

d: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1\)

\(\Leftrightarrow10x^2-19x=0\)

=>x(10x-19)=0

=>x=0 hoặc x=19/10

Theo bài ra ta có : \(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\)

\(5y=7z\Rightarrow\dfrac{y}{7}=\dfrac{z}{5}\)

Lại có : \(\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\) (1)

\(\dfrac{y}{7}=\dfrac{z}{5}\Rightarrow\dfrac{y}{14}=\dfrac{z}{10}\) (2)

Từ (1) và (2) suy ra: \(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)

\(\Rightarrow\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}\) và \(3x-7y+5z=-30\)

Áp dụng tính chất dáy tỉ số bằng nhau ta có :

\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{-30}{15}=-2\)

\(\dfrac{3x}{63}=-2\Rightarrow3x=-126\Rightarrow x=-42\)

\(\dfrac{7y}{98}=-2\Rightarrow7y=-196\Rightarrow y=-28\)

\(\dfrac{5z}{50}=-2\Rightarrow5z=-100\Rightarrow z=-20\)

Vậy \(x,y,z\) lần lượt là \(\left(-42\right),\left(-28\right)\) và \(\left(-20\right)\)

a) 

5.(12-x)-20=30

⇒60-5x-20=30

⇒-5x=30+20-60

⇒-5x=-10

⇒x=2

b)(17x - 25 ) : 8 + 65 = 9²

(17x - 25 ) : 8 + 65 = 81

17x - 25 = 16 x 8 = 128

17x = 128+25=153

x= 153:17 =9

c)

x=23

Giải thích các bước giải:

3x – 10 = 2x + 13

3x-2x=13+10

x=23

d)4(2x+7)-3(3x-2)=24

4.2x+4.7-3.3x+3.2=24

8x+28-9x+6=24

8x-9x=24-28-6=-10

=>(-1)x=-10

        x=-10:(-1)

        x=10

a. \(5\cdot\left(12-x\right)-20=30\Leftrightarrow5\left(12-x\right)=50\)

\(\Leftrightarrow12-x=50:5=10\)

\(\Leftrightarrow x=12-10=2\)

b. \(\left(17x-25\right):8+65=9^2\)

\(\Leftrightarrow\left(17x-25\right):8=81-65=16\)

\(\Leftrightarrow17x-25=16:8=2\)

\(\Leftrightarrow17x=2+25=27\Leftrightarrow x=\frac{27}{17}\)

c. \(3x-10=2x+13\)

\(\Leftrightarrow3x-2x=10+13\)

\(\Leftrightarrow x=23\)

d. \(4\cdot\left(2x+7\right)-3\cdot\left(3x-2\right)=24\)

\(\Leftrightarrow8x+28-9x+6=24\)

\(\Leftrightarrow34-x=24\Leftrightarrow x=10\)

a) 

5.(12-x)-20=30

⇒60-5x-20=30

⇒-5x=30+20-60

⇒-5x=-10

⇒x=2

b)(17x - 25 ) : 8 + 65 = 9²

(17x - 25 ) : 8 + 65 = 81

17x - 25 = 16 x 8 = 128

17x = 128+25=153

x= 153:17 =9

c)

x=23

Giải thích các bước giải:

3x – 10 = 2x + 13

3x-2x=13+10

x=23

d)4(2x+7)-3(3x-2)=24

4.2x+4.7-3.3x+3.2=24

8x+28-9x+6=24

8x-9x=24-28-6=-10

=>(-1)x=-10

        x=-10:(-1)

        x=10

có gì ko hiểu bạn hỏi nhé

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

Ta có:

\(2x+1=0\Leftrightarrow x=\frac{-1}{2}\)

\(x-1=0\Leftrightarrow x=1\)

Lập bảng xét dấu :

+) Với  \(x< \frac{-1}{2}\Rightarrow\hept{\begin{cases}2x+1< 0\\x-1< 0\end{cases}\Rightarrow}\hept{\begin{cases}|2x+1|=-2x-1\\|x-1|=1-x\end{cases}\left(2\right)}\)

Thay (2) vào (1) ta được :

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

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

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

\(-4x=2\)

\(x=\frac{-1}{2}\)( loại ) 

+)  Với \(\frac{-1}{2}\le x< 1\Rightarrow\hept{\begin{cases}2x+1>0\\x-1< 0\end{cases}\Rightarrow\hept{\begin{cases}|2x+1|=2x+1\\|x-1|=1-x\end{cases}\left(3\right)}}\)

Thay (3) vào (1) ta được :

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

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

\(3x=3x\)( luôn đúng chọn )

+) Với \(x\ge1\Rightarrow\hept{\begin{cases}2x+1>0\\x-1>0\end{cases}\Rightarrow\hept{\begin{cases}|2x+1|=2x+1\\|x-1|=x-1\end{cases}\left(4\right)}}\)

Thay (4) vào (1) ta được :

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

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

\(2x-x-3x=-1-1\)

\(-2x=-2\)

\(x=1\)( chọn )

Vậy \(\frac{-1}{2}\le x\le1\)

\(\left|2x+1\right|-\left|x-1\right|=3x\Rightarrow\left|2x+1-1+x\right|\ge3x\)

\(\Leftrightarrow\left|3x\right|\ge3x\Rightarrow x\in\left\{x\inℤ|x\le0\right\}\)

`tan (1/2) ≈ 26^o 33'`

😭😭😭

ko ai quên đc đâu

còn lâu mới quên đc

\(\frac{2x}{25}=3x+\frac{10}{50}\)

\(\Leftrightarrow\frac{2x}{25}=\frac{75x}{25}+\frac{5}{25}\)

\(\Leftrightarrow2x=75x+5\)

\(\Leftrightarrow-73x=5\)

\(\Leftrightarrow x=-\frac{5}{73}\)

CHO MÌNH XIN

25 ko dùng in spite of mà dùng because of nhé

Because of being careless in writing, he got bad marks

26 In spite of being ill, she insisted on going to work

27 In spite of my trying to persuade her, i didn't succeed

\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)

\(=\left(\dfrac{2x+1}{-\left(2x+1\right)}-\dfrac{2x-1}{-\left(2x-1\right)}\right):\dfrac{4x}{10x-5}\)

\(=\dfrac{2x}{5x-5}\)