a: \(x\left(1-2x\right)+2x^2=14\)

=>\(x-2x^2+2x^2=14\)

=>x=14

b: \(x\left(x-5\right)+3x-15=0\)

=>\(\left(x-5\right)\left(x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

a) \(=x^2+7x-12x-84-2x+14\)

\(=x^2-7x-70\)

b)\(=x^2-4x-2x+8\)

\(=x\left(x-4\right)-2\left(x-4\right)\)

 \(=\left(x-4\right)\left(x-2\right)\)

c) \(=9x\left(x+y\right)-\left(x+y\right)\)

\(=\left(9x-1\right)\left(x+y\right)\)

d)\(=\left(x-y\right)^2-9^2\)

\(=\left(x-y+9\right)\left(x-y-9\right)\)

e)\(=x^2+8x+16-60+15x\)

\(=x^2+23x-44\)

a) Ta có: \(\left(x-12\right)\left(x+7\right)-2x-14\)

\(=\left(x-12\right)\left(x+7\right)-2\left(x+7\right)\)

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

b) Ta có: \(x^2-6x+8\)

\(=x^2-2x-4x+8\)

\(=\left(x-2\right)\left(x-4\right)\)

c) Ta có: \(9x^2+9xy-\left(x+y\right)\)

\(=9x\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(9x-1\right)\)

d) Ta có: \(\left(x^2-2xy+y^2\right)-81\)

\(=\left(x-y\right)^2-81\)

\(=\left(x-y-9\right)\left(x-y+9\right)\)

a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)

b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)

Bài 1: 

a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)

\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)

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

b) Ta có: \(\left(2x-3\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)

\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)

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

Bài 2: 

a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)

b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)

c) \(3+3^2+3^3+...+3^{2007}\)

\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{2005}\right)⋮13\)

a) x : 2 = y : (-5)

⇒ x/2 = y/(-5)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

x/2 = y/(-5) = (x - y)/(2 + 5) = 14/7 = 

x/2 = 2 ⇒ x = 2.2 = 4

y/(-5) = 2 ⇒ y = 2.(-5) = -10

Vậy x = 4; y = -10

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

x/2 = y/5 = z/6 = (x - y + z)/(2 - 5 + 6) = 24/3 = 8

x/2 = 8 ⇒ x = 8.2 = 16

y/5 = 9 ⇒ y = 8.5 = 40

z/6 = 8 ⇒ z = 8.6 = 48

Vậy x = 16; y = 40; z = 48

c) 2x = 3y = 6z

⇒ x/(1/2) = y/(1/3) = z/(1/6)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

x/(1/2) = y/(1/3) = z/(1/6) = (x + y - z)/(1/2 + 1/3 - 1/6) = 8/(2/3) = 12

2x = 12 ⇒ x = 12 : 2 = 6

3y = 12 ⇒ y = 12 : 3 = 4

6z = 12 ⇒ z = 12 : 6 = 2

Vậy x = 6; y = 4; z = 2

d) x/3 = y/2 = z/(-3)

⇒ 2x/6 = 3y/6 = 4z/(-12)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

2x/6 = 3y/6 = 4z/(-12) = (2x - 3y + 4z)/(6 - 6 - 12) = 48/(-12) = -4

x/3 = -4 ⇒ x = -4.3 = -12

y/2 = -4 ⇒ y = -4.2 = -8

z/(-3) = -4 ⇒ z = -4.(-3) = 12

Vậy x = -12; y = -8; z = 12

e) Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

x/5 = y/6 = z/7 = (x - y)/(5 - 6) = 36/(-1) = -36

x/5 = -36 ⇒ x = -36.5 = -180

y/6 = -36 ⇒ y = -36.6 = -216

z/7 = -36 ⇒ z = -36.7 = -252

Vậy x = -180; y = -216; z = -252

f) x/12 = y/13

⇒ 3x/36 = 2y/26

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

3x/36 = 2y/26 = (3x + 2y)/(36 + 26) = 52/62 = 26/31

x/12 = 26/31 ⇒ x = 26/31 . 12 = 312/31

y/13 = 26/31 ⇒ y = 26/31 . 13 = 338/31

z/15 = 26/31 ⇒  z = 26/31 . 15 = 390/31

Vậy x = 312/31; y = 338/31; z = 390/31

c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x=0\)

\(\Leftrightarrow x\left(3x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)

\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

A

A

A nha 

Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(5\cdot x^3-5=0\)

`=> 5*x^3 = 0+5`

`=> 5*x^3 = 5`

`=> x^3 = 5 \div 5`

`=> x^3 = 1`

`=> x^3 = 1^3`

`=> x=1`

Vậy, `x=1.`

`b)`

\(( x+1)^2 = 16\)

`=> (x+1)^2 = (+-4)^2`

`=>`\(\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4-1\\x=-4-1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy, `x \in {3; -5}`

`c)`

\(( x+1)^3 = 27\)

`=> (x+1)^3 = 3^3`

`=> x+1=3`

`=> x=3-1`

`=> x=2`

Vậy, `x=2.`

`d)`

\(( x-1)^3 = 343\)

`=> (x-1)^3 = 7^3`

`=> x-1=7`

`=> x=7+1`

`=> x=8`

Vậy, `x=8.`

`e)`

\((2x - 1^3) = 125\) hay đề là `(2x-1)^3 = 125` vậy ạ?

Mình làm cả 2 TH nhé!

`(2x-1^3)=125`

`=> 2x-1=125`

`=> 2x=125+1`

`=> 2x=126`

`=> x=126 \div 2`

`=> x=63`

TH2:

`(2x-1)^3 = 125`

`=> (2x-1)^3 = 5^3`

`=> 2x-1=5`

`=> 2x=5+1`

`=> 2x=6`

`=> x=6 \div 2`

`=> x=3`

Vậy, `x=3.`

(a) \(5x^3-5=0\Leftrightarrow5x^3=5\Leftrightarrow x^3=1\Leftrightarrow x=1\)

(b) \(\left(x+1\right)^2=16\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

(c) \(\left(x+1\right)^3=27\Leftrightarrow x+1=3\Leftrightarrow x=2\)

(d) \(\left(x-1\right)^3=343\Leftrightarrow x-1=7\Leftrightarrow x=8\)

(e) \(\left(2x-1\right)^3=125\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)

a) x=3 y=13

x=16 y=0

x=4 y=5

x=9 y=1