Tìm x , biết
a. ( x-3)\(^2\)-4 = 0
b.( x+4)\(^2\)- (x+1).(x-1)=16
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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) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
PN
2
15 tháng 8 2021
a: Ta có: \(\left(x+1\right)^2-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b: Ta có: \(2\left(3x-2\right)^2=9x^2-4\)
\(\Leftrightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(6x-4-3x-2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
28 tháng 7 2023
2:
a: =>(x-9)(x-1)=0
=>x=9 hoặc x=1
b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0
=>(x+4)(x^2-4x+16+x-16)=0
=>(x+4)(x^2-3x)=0
=>x(x-3)(x+4)=0
=>x=0;x=3;x=-4
YN
28 tháng 7 2023
bài 2 :
a: =>(x-9)(x-1)=0
=>x=9 hoặc x=1
b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0
=>(x+4)(x^2-4x+16+x-16)=0
=>(x+4)(x^2-3x)=0
=>x(x-3)(x+4)=0
=>x=0;x=3;x=-4
D
1
25 tháng 10 2021
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(x^2-1=0\Rightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
25 tháng 7 2021
a) (x-2)3+6(x+1)2-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0
\(\Rightarrow\)24x+10=0
\(\Rightarrow\)24x=-10
\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
25 tháng 7 2021
b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2
\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2
\(\Rightarrow\)x2-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2
\(\Rightarrow\)3x2-18x-22=3x2+2x+17
\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0
\(\Rightarrow\)-20x-39=0
\(\Rightarrow\)-20x=39
\(\Rightarrow\)x=\(-\dfrac{39}{20}\)
NA
1
SN
20 tháng 7 2021
a. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
16 tháng 9 2023
\(Bài.1:\\ a,104^2-16=104^2-4^2=\left(104+4\right)\left(104-4\right)=108.100=10800\\ b,9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\\ =\left(9.2\right)^8-\left(18^8-1\right)=18^8-18^8+1=1\\ c,999^3+3.999^2+3.999+1\\ =999^3+3.999^2.1+3.999.1^2+1^3=\left(999+1\right)^3=1000^3=1000000000\\ d,42^3-6.42^2+12.42-8\\ =42^3-3.42^2.2+3.42.2^2-2^3\\ =\left(42-2\right)^3=40^3=64000\)
KV
16 tháng 9 2023
Bài 1
a) 104² - 16
= 104² - 4²
= (104 - 4)(104 + 4)
= 100.108
= 10800
b) 9⁸.2⁸ - (18⁴ - 1)(18⁴ + 1)
= 18⁸ - (18⁸ - 1)
= 18⁸ - 18⁸ + 1
= 1
c) 999³ + 3.999² + 3.999 + 1
= (999 + 1)³
= 1000³
= 1000000000
d) 42³ - 6.42² + 12.42 - 8
= (42 - 2)³
= 40³
= 64000
21 tháng 6 2023
a: (x+1)^3-x(x-2)^2+x-1=0
=>x^3+3x^2+3x+1-x(x^2-4x+4)+x-1=0
=>x^3+3x^2+4x-x^3+4x^2-4x=0
=>7x^2=0
=>x=0
b: =>x^3-3x^2+3x-1-x^3-27+3x^2-12=2
=>3x=2+1+27+12=39+3=42
=>x=14
14 tháng 11 2021
a) (x-4)(x+4)-x(x+2)=0
x2-16-x2-2x = 0
-16 - 2x = 0
2x = -16
x = -16/2
x = -8
b) 3x(x-2)-x+2=0
(3x-1)(x-2)=0
=> x ∈ {1/3 ; 2 }
c) 6x - 12x2 = 0
6x(1-2x) = 0
=> x ∈ {0; 1/2 }
d) mình thấy có vẻ hơi sai đề nên mình ko giải được, bạn thông cảm nha
17 tháng 8 2021
a: Ta có: \(4\left(x+1\right)^2+\left(2x+1\right)^2-8\left(x-1\right)\left(x+1\right)-11=0\)
\(\Leftrightarrow4x^2+8x+4+4x^2+4x+1-8x^2+8-11=0\)
\(\Leftrightarrow12x=-2\)
hay \(x=-\dfrac{1}{6}\)
b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)-1=0\)
\(\Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\)
\(\Leftrightarrow2x=-40\)
hay x=-20
a) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=4\\x-3=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)
b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x+17=16\)
\(\Leftrightarrow8x=-1\)
\(\Leftrightarrow x=\frac{-1}{8}\)