TIM X BIET:A/ (X-3).(X-1/2)=0B/ X2-2X=0C/(3X-1).(X2 1)=0D/ (X-2).(X 1)=0

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

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

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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}

DC

Đinh Cẩm Tú

14 tháng 1 2021

Giải các phương trình:

a) (x - 7)(2x + 8) = 0

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

c) (x - 1)(2x + 7)(x² + 2) = 0

d) (2x - 1)(x + 8)(x - 5) = 0

7

TG

Trúc Giang

14 tháng 1 2021

a) (x - 7)(2x + 8) = 0

$\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=7\\2x=-8\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.$

Vậy: S = {7; -4}

b) Tương tự câu a

c) (x - 1)(2x + 7)(x² + 2) = 0

$\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\\x^2+2=0\end{matrix}\right.$

Mà: x²+ 2 > 0 với mọi x

$\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-7\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.$

Vậy: $S=\left\{1;-\dfrac{7}{2}\right\}$

d) (2x - 1)(x + 8)(x - 5) = 0

$\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-8\\x=5\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.$

Vậy $S=\left\{\dfrac{1}{2};-8;5\right\}$

TT

Thu Thao

14 tháng 1 2021

a/ Pt $\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.$

Vậy $S=\left\{7;-4\right\}$

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

c/ pt $\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.$ ($x^2+2>0\forall x$)$\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.$

d/ pt $\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.$

QT

Quynh Tram Nguyenn

2 tháng 1 2022

Tìm x

a, 3/4x*(x²-9)=0

b, x³-16x=0

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

d, 3x³-27x=0

e, x²(x+1)+2x(x+1)=0

f, x(2x-3)-2(3-2x)=0

3

NL

Nguyễn Lê Phước Thịnh

2 tháng 1 2022

c: =>(x-1)(x+1)=0

hay $x\in\left\{1;-1\right\}$

QT

Quynh Tram Nguyenn

2 tháng 1 2022

plss

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: $2x^2+7x+3=0$

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

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

CI

Cíu iem

21 tháng 9 2021

Tìm x:
a) x(x-2)-x+2=0
b) (3-2x)²-(x-1)²=0
c) 81x⁴-x²=0
d) x³+x²+27x+27=0

2

RH

Rin Huỳnh

21 tháng 9 2021

a) (x-1)(x-2)=0

x-1=0 --> x = 1

x-2=0 --> x = 2

RH

Rin Huỳnh

21 tháng 9 2021

d) x^2(x + 1) + 27(x + 1)=0

(x+1)(x^2+27)=0

x+1=0 --> x = -1

x^2+27=0 (vô lí)

CI

Cíu iem

21 tháng 9 2021

Tìm x:
a) x(x-2)-x+2=0
b) (3-2x)²-(x-1)²=0
c) 81x⁴-x²=0
d) x³+x²+27x+27=0

4

MH

Minh Hiếu

21 tháng 9 2021

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

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

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

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

PA

Phan An

21 tháng 9 2021

a) $x (x - 2) - x + 2 = 0$

$x (x - 2) - (x - 2) = 0$

$(x - 1) (x - 2) = 0$

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

H

Hiếu

2 tháng 11 2021

Tìm x biết:

1,

a,3x(x+1) - 2x(x+2) = -x-1

b,2x(x-2020) - x+2020 = 0

c,(x-4)² - 36 = 0

d,x²+ 8x - 16 = 0

e,x(x+6) - 7x - 42 = 0

f,25x² - 16 = 0

2,

a,3x³ - 12x = 0

b,x² + 3x - 10 = 0

1

LL

Lấp La Lấp Lánh

2 tháng 11 2021

Bài 1:

a) $\Rightarrow3x^2+3x-2x^2-4x+x+1=0$

$\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing$

b) $\Rightarrow\left(x-2020\right)\left(2x-1\right)=0$

$\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.$

c) $\Rightarrow\left(x-10\right)\left(x+2\right)=0$

$\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.$

d) $\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4$

e) $\Rightarrow\left(x+6\right)\left(x-7\right)=0$

$\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.$

f) $\Rightarrow\left(5x-4\right)\left(5x+4\right)=0$

$\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.$

Bài 2:

a) $\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0$

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

b) $\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0$

$\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.$

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NL

Nguyễn Linh

3 tháng 2 2022

Giải các phương trình sau:

a/ (3x – 2)(4x + 5) = 0
b/ (2,3x – 6,9)(0,1x + 2) = 0
c/ (4x + 2)(x² + 1) = 0
d/(2x + 7)(x – 5)(5x + 1) = 0
e/ (x – 1)(2x + 7)(x² + 2) = 0
f/ (3x + 2)(x² – 1) = (9x² – 4)(x + 1)

1

NT

Nguyễn Thái Thịnh

3 tháng 2 2022

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

$\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.$

Vậy: $S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}$

b) $\left(2,3x-6,9\right)\left(0,1x+2\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.$

c) $\left(4x+2\right)\left(x^2+1\right)=0$

Vì $x^2+1\ge1>0\forall x$

$\Rightarrow4x+2=0$

$\Leftrightarrow x=-\dfrac{1}{2}$

Vậy: $S=\left\{-\dfrac{1}{2}\right\}$

d) $\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.$

Vậy: $S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}$

e) $\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0$

Vì $x^2+2\ge2>0\forall x$

$\Rightarrow\left(x-1\right)\left(2x+7\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.$

f) $\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)$

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

$\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0$

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

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

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

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

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

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

Vậy: $S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}$

D

đăng2k7:)))

11 tháng 6 2021

câu 13 trong các phương trình sau phương bậc nhất một ẩn là

a) x+x2=0

b) 1/x+1=0

c) 1/2x-2=0

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