Tìm x biết:a) (3x 5)2 - 4x2 = 0b) 25x4 - (4x - 3)2 = 0c) (3x 7)2 - (2x - 3)2 = 0d) (4x - 1)2 - (5 - 3x)2 = 0

1

NL

Nguyễn Lê Phước Thịnh

19 tháng 8 2021

a: Ta có: $\left(3x+5\right)^2-4x^2=0$

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

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

$\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.$

TL

Tuyết Ly

23 tháng 10 2021

Tìm x biết:

a, 16x² – 9(x + 1)²= 0

b, x2 (x – 1) – 4x2 + 8x – 4 = 0

c, x(2x – 3) – 2(3 – 2x) = 0

d, (x – 3)(x² + 3x + 9) – x(x + 2)(x – 2) = 1

e, 4x² + 4x – 6 = 2

f, 2x² + 7x + 3 = 0

1

NL

Nguyễn Lê Phước Thịnh

23 tháng 10 2021

e: ta có: $4x^2+4x-6=2$

$\Leftrightarrow4x^2+4x-8=0$

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

$\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.$

f: Ta có: $2x^2+7x+3=0$

$\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.$

Đúng(2)

MT

Minh Trí Nguyễn

9 tháng 12 2021

tìm x, biết:

a) 9x²+36=0

b) 3(x+4)-x²-4x=0

c) x(2x-1)-(x-2)(2x+1)=0

d) (2x-3)²-4x²=00

e)1 phần 3.x²-3x=0

f) x³-x²-x+1=0

ráng giúp mình nha

1

NH

Nguyễn Hoàng Minh

9 tháng 12 2021

$a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.$

NM

Nguyen Minh Anh

13 tháng 11 2021

Bài 5. Tìm x, biết:

a) x (2x - 7) + 4x -14 = 0

b) x³ - 9x = 0

c) 4x² -1 - 2(2x -1)2 = 0

d) (x³ - x² ) - 4x² + 8x - 4 = 0

1

MD

Monkey D. Luffy

13 tháng 11 2021

$a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.$

Đúng(3)

DC

Đinh Cẩm Tú

11 tháng 1 2021

Tìm x, biết:

a) 7x² - 28 = 0

b) $\dfrac{2}{3}$x(x² - 4) = 0

c) 2x(3x - 5) - (5 - 3x) = 0

d) (2x - 1)² - 25 = 0

3

NL

Nguyễn Lê Phước Thịnh

11 tháng 1 2021

a) Ta có: $7x^2-28=0$

$\Leftrightarrow7\left(x^2-4\right)=0$

$\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0$

mà 7>0

nên (x-2)(x+2)=0

hay $\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.$

Vậy: $x\in\left\{2;-2\right\}$

b) Ta có: $\dfrac{2}{3}x\left(x^2-4\right)=0$

$\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0$

mà $\dfrac{2}{3}>0$

nên x(x-2)(x+2)=0

hay $\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.$

Vậy: $x\in\left\{0;-2;2\right\}$

c) Ta có: $2x\left(3x-5\right)-\left(5-3x\right)=0$

$\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0$

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

$\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.$

Vậy: $x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}$

d) Ta có: $\left(2x-1\right)^2-25=0$

$\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0$

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

$\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.$

Vậy: $x\in\left\{3;-2\right\}$

MC

Minh Cao

11 tháng 1 2021

a,7x² - 28 = 0

=> 7x² = 28 => x² = 4 => x = 2

b,2/3x(x² - 4) = 0

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

=> x ∈ {0 ; 2 ; -2}

c,2x(3x - 5) - (5 - 3x) = 0

= 2x(3x - 5) + (3x - 5)

= (3x - 5)(2x + 1) = 0

=> x ∈ { 5/3 ; -1/2}

d, (2x - 1)² - 25 = 0

=> (2x - 4)(2x - 6) = 0

=> x ∈ {2 ;3}

C

Cíuuuuuuuuuu

24 tháng 8 2021

Tìm x:
a) x⁴-25x³=0
b) (x-5)²-(3x-2)²=0
c) x³-4x²-9x+36=0
d) (-x³+3x²-4x) : ($-\dfrac{1}{2}$x)=0

6

AH

Akai Haruma

Giáo viên

24 tháng 8 2021

a.

$x^4-25x^3=0$

$\Leftrightarrow x^3(x-25)=0$

$\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.$

b.

$(x-5)^2-(3x-2)^2=0$

$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$

$\Leftrightarrow (-2x-3)(4x-7)=0$
$\Leftrightarrow \left[\begin{matrix} -2x-3=0\\ 4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{7}{4}\end{matrix}\right.$

AH

Akai Haruma

Giáo viên

24 tháng 8 2021

c.

$x^3-4x^2-9x+36=0$

$\Leftrightarrow x^2(x-4)-9(x-4)=0$

$\Leftrightarrow (x-4)(x^2-9)=0$

$\Leftrightarrow (x-4)(x-3)(x+3)=0$

$\Leftrightarrow \left[\begin{matrix} x-4=0\\ x-3=0\\ x+3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=3\\ x=-3\end{matrix}\right.$

d. ĐK: $x\neq 0$

$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$

$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$

$\Leftrightarrow -2(-x^2+3x-4)=0$

$\Leftrightarrow x^2-3x+4=0$

$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)

Vậy pt vô nghiệm.

KH

khánh huyền

30 tháng 7 2021

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0
b) 3x (x-1) + x - 1 = 0
c) 2(x+3) - x ² - 3x = 0
d) x(x - 2) + 3x - 6 = 0
e) 4x ² - 4x +1 = 0
f) x +5x ² = 0
g) x ² 2x -3 = 0

4

PS

Phía sau một cô gái

30 tháng 7 2021

a) $\left(2x-1\right)^2-25=0$

⇔ $\left(2x-1\right)^2-5^2=0$

⇔ $\left(2x-1-5\right)\left(2x-1+5\right)=0$

⇒ $2x-1-5=0$ hoặc $2x-1+5=0$

⇔ $x=3$ hoặc $x=-2$

TC

Thùy Cái

30 tháng 7 2021

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0

<=> (2x-1)² = 25

<=> 2x-1 = 5 hay 2x-1 =-5

<=> 2x= 6 hay 2x=-4

<=> x=3 hay x= -2

Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0

<=> (x-1)(3x+1)=0

<=> x-1=0 hay 3x+1=0

<=> x=1 hay 3x=-1

<=> x=1 hay x=$\dfrac{-1}{3}$

Vậy S={1;$\dfrac{-1}{3}$}

c) 2(x+3) - x ² - 3x = 0

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

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

<=> x+3=0 hay 2-x=0

<=> x=-3 hay x=2

Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0

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

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

<=> x-2=0 hay x+3=0

<=> x=2 hay x=-3

Vậy S={2;-3}
e) 4x ² - 4x +1 = 0

<=> (2x-1)²=0

<=> 2x-1=0

<=> 2x=1

<=> x=$\dfrac{1}{2}$

Vậy S={$\dfrac{1}{2}$}
f) x +5x² = 0

<=> x(1+5x)=0

<=>x=0 hay 1+5x=0

<=> x=0 hay 5x=-1

<=> x=0 hay x= $\dfrac{-1}{5}$

Vậy S={0;$\dfrac{-1}{5}$}
g) x ²+ 2x -3 = 0

<=> x²-x+3x-3=0

<=> x(x-1)+3(x-1)=0

<=> (x-1)(x+3)=0

<=> x-1=0 hay x+3=0

<=> x=1 hay x=-3

Vậy S={1;-3}

TV

Thư Vũ

14 tháng 11 2021

Bài 2. Tìm x, biết:

a/ (x – 4)(x + 4) - x(x + 2) = 0

b/ 3x(x – 2) – x + 2 = 0

c/ 6x - 12x2 = 0

d/ 4x(3 - 14x) + (x – 2)(x + 2) = 0

1

NH

Nguyễn Hoàng Minh

14 tháng 11 2021

$a,\Leftrightarrow x^2-16-x^2-2x=0\\ \Leftrightarrow2x=-16\Leftrightarrow x=-8\\ b,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow6x\left(1-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\\ d,\Leftrightarrow12x-56x^2+x^2-16=0\\ \Leftrightarrow55x^2-12x+16=0\\ \Delta=144-4\cdot55\cdot16< 0\\ \Leftrightarrow x\in\varnothing$

Đúng(3)

C

Cíuuuuuuuuuu

7 tháng 8 2021

Tìm x:
a) 36x³-4x=0
b) 3x(x-2)-2+x=0
c) (x³-x²)-4x²+8x-4=0
d) x²-6x-16=0
e) x⁴-6x²-7=0

1

TG

Trúc Giang

7 tháng 8 2021

undefined

VN

vân nguyễn

30 tháng 6 2021

Tìm x

a) 3x(4x - 3) - 2x(5 - 6x) = 0

b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0

c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)

d) 3x (x + 1) - 5x(3 - x) + 6(x^2 + 2x + 3) = 0

3

K

Kenny

30 tháng 6 2021

a) 3x(4x-3)-2x(5-6x)=0

$\Leftrightarrow12x^2-9x-10x+12x^2=0$

$\Leftrightarrow24x^2-19x=0$

$\Leftrightarrow x\left(24x-19\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.$

Vậy x=0 hoặc x=$\dfrac{19}{24}$

K

Kenny

30 tháng 6 2021

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

$\Leftrightarrow$10x-15+4x²-8x+6x-4x²=0

$\Leftrightarrow8x-15=0$

$\Leftrightarrow8x=15$

$\Leftrightarrow x=\dfrac{15}{8}$

vậy x=$\dfrac{15}{8}$