1, Tìm x, biết:
A) 2x(x-7)+5x-35=0
B) x(x-3)-7x+21=0
C) x^3-x^2-25x+25=0
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
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: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
tìm x bt
a)x(x-5)-4x+20=0
b)x(x+6)-7x-42=0
c)x^3-5x^2-x+5=0
d)4x^2-25-(2x-5)(3x+7)=0
e)x^3+27+(x+3)(x-9)=0
giúp mk vs ah!!1
a) x(x - 5) - 4x + 20 = 0
\(\Leftrightarrow\) x(x - 5) - (4x + 20)
\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x - 4)
Khi x - 5 = 0 hoặc x - 4 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 4
Vậy S = \(\left\{5;4\right\}\)
b) x(x + 6) - 7x - 42 = 0
\(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0
\(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0
\(\Leftrightarrow\) (x + 6)(x - 7) = 0
Khi x - 6 = 0 hoặc x - 7 = 0
\(\Leftrightarrow\) x = 6 \(\Leftrightarrow\) x = 7
Vậy S = \(\left\{6;7\right\}\)
c) x3 - 5x2 - x + 5 = 0
\(\Leftrightarrow\) (x3 - 5x2) - (x + 5) = 0
\(\Leftrightarrow\) x2 (x - 5) - (x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x2 - 1) = 0
\(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0
Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 1 \(\Leftrightarrow\) x = -1
Vậy S = \(\left\{5;1;-1\right\}\)
d) 4x2 - 25 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x)2 - 52 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0
\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0
Khi 2x - 5 = 0 hoặc -x + 12 = 0
\(\Leftrightarrow\) 2x = 5 \(\Leftrightarrow\) -x = -12
\(\Leftrightarrow\) x = \(\dfrac{5}{2}\) \(\Leftrightarrow\) x = 12
Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)
e) x3 + 27 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) x3 - 33 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 3x + 9) + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0
\(\Leftrightarrow\) (x - 3) ( x2 - 3x + 9 + x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 2x) = 0
\(\Leftrightarrow\) (x - 3)x(x - 2)
Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x = 3 \(\Leftrightarrow\) x = 2
Vậy S = \(\left\{3;0;2\right\}\)
Chúc bạn học tốt
a) Ta có: \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
b) Ta có: \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
c) Ta có: \(x^3-5x^2-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)
d) Ta có: \(4x^2-25-\left(2x-5\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-3x-7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
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
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: \(x\in\left\{3;-2\right\}\)
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}
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}
Tìm x,biết:
a)(2x-3)2-49=0
b)2x.(x-5)-7.(5-x)=0
c)x2-3x-10=0
a) \(\Rightarrow\left(2x-3\right)^2=49\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c) \(\Rightarrow x\left(x-5\right)+2\left(x-5\right)=0\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a, ⇒ (2x - 3)2 = 49
⇒ (2x - 3)2 = \(\left(\pm7\right)^2\)
⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0
⇒ (x - 5).(2x + 7) = 0
⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c, ⇒ x2 - 5x + 2x - 10 = 0
⇒ (x2 - 5x) + (2x - 10) = 0
⇒ x.(x - 5) +2.(x - 5) = 0
⇒ (x - 5).(x + 2)=0
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Bài 7. Tìm x,biết:
a) x-3x2=0 e) 5x(3x-1)+x(3x-1)-2(3x-1)=0
b) (x+3)2-x(x-2)=13 c) (x-4)2-36=0
d) x2-7x+12=0 g) x2-2018x-2019=0
Bài 8. Tìm x, biết
a) (2x-1)2=(x+5)2 b) x2-x+1/4
c) 4x4-101x2+25=0 d) x3-3x2+9x-91=0
Tìm x biết:
a) 7x.(2x - 3) - (4x2 - 9) = 0
b) (2x - 7).(x - 2).(x2 - 4) = 0
c) (9x2 - 25) - (6x - 10) = 0
a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)
a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)
\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
Tìm các số thực x, biết:
a) (2x-3)2-49=0
b) 2x(x-5)-7(5-x)=0
c) x2-3x-10=0
a: \(\left(2x-3\right)^2-49=0\)
\(\Leftrightarrow\left(2x+4\right)\left(2x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
a. (2x - 3)2 - 49 = 0
<=> (2x - 3)2 - 72 = 0
<=> (2x - 3 + 7)(2x - 3 - 7) = 0
<=> (2x + 4)(2x - 10) = 0
<=> \(\left[{}\begin{matrix}2x+4=0\\2x-10=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
b. 2x(x - 5) - 7(5 - x) = 0
<=> 2x(x - 5) + 7(x - 5) = 0
<=> (2x + 7)(x - 5) = 0
<=> \(\left[{}\begin{matrix}2x+7=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
c. x2 - 3x - 10 = 0
<=> x2 - 5x + 2x - 10 = 0
<=> x(x - 5) + 2(x - 5) = 0
<=> (x + 2)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
a, (2x - 3)2 - 49 = 0
(2x - 3)2 - 72 = 0
(2x - 3 + 7)( 2x - 3 - 7) = 0
(2x + 4)( 2x - 10) = 0
=> 2x + 4 = 0 => 2x - 10 = 0
2x = - 4 2x = 10
x = - 2 x = 5
a: =>(x-5)(x+5)+(x-5)(3x-15)=0
=>(x-5)(x+5+3x-15)=0
=>(x-5)(4x-10)=0
=>x=5 hoặc x=5/2
c: =>x^3-3x^2+2x^2-6x-8x+24=0
=>(x-3)(x^2+2x-8)=0
=>(x-3)(x+4)(x-2)=0
=>\(x\in\left\{3;-4;2\right\}\)
Tìm tập hợp giá trị của x biết:
a,(x-1)(x-2)>0
b,2x-3<0
c,(2x-4)(9-3x)>0
d,2x/3-3/4<0