Tìm x:
a) ( x + 5 )2 - x2 = 45
b) ( 2x - 3 )2 - ( 2x + 1 )2 = 3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
30 tháng 9 2021
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
4 tháng 9 2021
a: Ta có: \(2x^3-18x=0\)
\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)
\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)
\(\Leftrightarrow-13x=13\)
hay x=-1
c: Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8=3-3x^2\)
\(\Leftrightarrow3x=12\)
hay x=4
4 tháng 9 2021
a) 2x3-18x=0
⇔ 2x(x2-9)=0
⇔ 2x(x-3)(x+3)=0
⇔ \(\left\{{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b)(3x-1)(2x+1)-6x(x+2)=11
⇔ 6x2+x-1-6x2-12x=11
⇔ -11x=12
\(\Leftrightarrow x=-\dfrac{12}{11}\)
c) (x-1)3-(x+2).(x2-2x+4)=3.(1-x2)
⇔ x3-3x2+3x-1-x3-8-3+3x2=0
⇔ 3x=12
⇔ x=4
T
2
KV
31 tháng 10 2023
a) (5 + x)(x - 5) - x(x + 5) = 10
x² - 25 - x² - 5x = 10
-5x = 10 + 25
-5x = 35
x = 35 : (-5)
x = -7
b) x.(2x + 3) - 2(x² + x) = 2
2x² + 3x - 2x² - 2x = 2
x = 2
31 tháng 10 2023
a: \(\left(x+5\right)\left(x-5\right)-x\left(x+5\right)=10\)
=>\(x^2-25-x^2-5x=10\)
=>-5x-25=10
=>-5x=35
=>x=-7
b: \(x\left(2x+3\right)-2\left(x^2+x\right)=2\)
=>\(2x^2+3x-2x^2-2x=2\)
=>x=2
MT
2
14 tháng 12 2021
\(a,\Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\\x=-3\end{matrix}\right.\\ b,\Leftrightarrow\left[{}\begin{matrix}1-2x=3x-2\\2x-1=3x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)
14 tháng 12 2021
a) \(\Leftrightarrow x^2\left(x+5\right)-9x-45=0\)
\(\Leftrightarrow x^2\left(x+5\right)-9x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\\x=-3\end{matrix}\right.\)
Vậy...
b) \(\Leftrightarrow\left(1-2x\right)^2-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\)
\(\Leftrightarrow\left(3-5x\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-5x=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)
Vậy...
C
2
21 tháng 8 2021
a: Ta có: \(\left(2x-3\right)^2+6\left(2x-1\right)=7\)
\(\Leftrightarrow\left(2x-3\right)^2+6\left(2x-1\right)-7=0\)
\(\Leftrightarrow4x^2-12x+9+12x-6-7=0\)
\(\Leftrightarrow4x^2=4\)
\(\Leftrightarrow x^2=1\)
hay \(x\in\left\{1;-1\right\}\)
b: Ta có: \(x^2-7x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
C
4
21 tháng 8 2021
a) \(\left(2x-3\right)^2+6\left(2x-1\right)=7\\ \Rightarrow4x^2-12x+9+12x-6-7=0\\ \Rightarrow4x^2-4=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
b) \(x^2-7x+10=0\\ \Rightarrow\left(x^2-2x\right)-\left(5x-10\right)=0\\ \Rightarrow\left(x-2\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
c) \(-6x^2+13x-5=0\\ \Rightarrow-\left(6x^2-13x+5\right)=0\\ \Rightarrow-\left[\left(6x^2-10x\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left[2x\left(3x-5\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left(2x-1\right)\left(3x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\left(2x-1\right)=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
28 tháng 8 2021
Lời giải:
a. PT $\Leftrightarrow (3-2x-3-2x)(3-2x+3+2x)=8$
$\Leftrightarrow -4x.6=8$
$\Leftrightarrow -24x=8\Leftrightarrow x=\frac{-1}{3}$
b.
$9x^5-72x^2=0$
$\Leftrightarrow 9x^2(x^3-8)=0$
$\Leftrightarrow x^2=0$ hoặc $x^3=8$
$\Leftrightarrow x=0$ hoặc $x=2$
c.
$5x^4-8x^2-4=0$
$\Leftrightarrow 5x^4-10x^2+2x^2-4=0$
$\Leftrightarrow 5x^2(x^2-2)+2(x^2-2)=0$
$\Leftrightarrow (5x^2+2)(x^2-2)=0$
$\Leftrightarrow 5x^2+2=0$ (loại) hoặc $x^2-2=0$ (chọn)
$\Leftrightarrow x=\pm \sqrt{2}$
d.
PT $\Leftrightarrow [x^2(x+1)-4(x+1)]:(x-2)=0$
$\Leftrightarrow (x^2-4)(x+1):(x-2)=0$
$\Leftrightarrow (x-2)(x+2)(x+1):(x-2)=0$
$\Leftrightarrow (x+2)(x+1)=0$
$\Leftrightarrow x+2=0$ hoặc $x+1=0$
$\Leftrightarrow x=-2$ hoặc $x=-1$
28 tháng 8 2021
a: Ta có: \(\left(3-2x\right)^2-\left(3+2x\right)^2=8\)
\(\Leftrightarrow9-12x+4x^2-9-12x-4x^2=8\)
\(\Leftrightarrow-24x=8\)
hay \(x=-\dfrac{1}{3}\)
b: Ta có: \(9x^5-72x^2=0\)
\(\Leftrightarrow9x^2\left(x^3-8\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
27 tháng 2 2021
`a,x(x-1)-(x+2)^2=1`
`<=>x^2-x-x^2-4x-4=1`
`<=>-5x=5`
`<=>x=-1`
`b,(x+5)(x-3)-(x-2)^2=-1`
`<=>x^2+2x-15-x^2+4x-4+1=0`
`<=>6x-18=0`
`<=>x-3=0`
`<=>x=3`
`c,x(2x-4)-(x-2)(2x+3)=0`
`<=>2x(x-2)-(x-2)(2x+3)=0`
`<=>(x-2)(2x-2x-3)=0`
`<=>-3(x-2)=0`
`<=>x-2=0`
`<=>x=2`
`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`
`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`
`<=>4x+26=-12`
`<=>4x=-38`
`<=>x=-19/2`
a) \(\left(x+5\right)^2-x^2=45\)
\(\Leftrightarrow\)\(\left(x+5-x\right)\left(x+5+x\right)=45\)
\(\Leftrightarrow\)\(5\left(x^2+5\right)=45\)
\(\Leftrightarrow\)\(x^2+5=\frac{45}{5}=9\)
\(\Leftrightarrow\)\(x^2=9-5=4\)
\(\Leftrightarrow\)\(x^2=2^2\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
b) \(\left(2x-3\right)^2-\left(2x+1\right)^2=3\)
\(\Leftrightarrow\)\(\left(2x-3-2x-1\right)\left(2x-3+2x+1\right)=3\)
\(\Leftrightarrow\)\(-4\left(4x-2\right)=3\)
\(\Leftrightarrow\)\(-8\left(2x-1\right)=3\)
\(\Leftrightarrow\)\(2x-1=-\frac{3}{8}\)
\(\Leftrightarrow\)\(2x=-\frac{3}{8}+\frac{8}{8}=\frac{5}{8}\)
\(\Leftrightarrow\)\(x=\frac{5}{16}\)
a) \(\left(x+5\right)^2-x^2=45\)
\(\Rightarrow x^2+2\times5x-x^2=45\)
\(\Rightarrow10x=45\)
\(\Rightarrow x=45\div10=4,5\)
Vậy giá trị của x là 4,5
b) \(\left(2x-3\right)^2-\left(2x+1\right)^2=3\)
\(\Rightarrow\left(2x-3+2x+1\right)\left(2x-3-2x-1\right)=3\)
\(\Rightarrow\left(4x-2\right)\left(-4\right)=3\)
\(\Rightarrow-16x+8=3\)
\(\Rightarrow-16x=-5\)
\(\Rightarrow x=\frac{5}{16}\)
Vậy giá trị của x là \(\frac{5}{16}\)