Tìm x :
a, 2640+(( 900 -3x )/45)*24=3000
b, 1404:(118-(4x+6))=27
c, 16/8x-2019/2021
d, 5xmũ 2 -3x=0
e, 3(x-1)+4(x-1)mũ 2 =0
f,2(x-2) mũ 2 =(x-2)
g, (x-2020) mũ 4 = ( x-2020) mũ 2
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F
5 tháng 10 2020
a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
F
5 tháng 10 2020
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)
24 tháng 9 2020
1) x3 - 3x2 = 0
<=> x2( x - 3 ) = 0
<=> \(\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
2) 5x( x - 2020 ) - x + 2020 = 0
<=> 5x( x - 2020 ) - ( x - 2020 ) = 0
<=> ( x - 2020 )( 5x - 1 ) = 0
<=> \(\orbr{\begin{cases}x-2020=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{1}{5}\end{cases}}\)
3) ( 3x - 5 )2 = ( x + 1 )2
<=> ( 3x - 5 )2 - ( x + 1 )2 = 0
<=> [ ( 3x - 5 ) - ( x + 1 ) ][ ( 3x - 5 ) + ( x + 1 ) ] = 0
<=> ( 3x - 5 - x - 1 )( 3x - 5 + x + 1 ) = 0
<=> ( 2x - 6 )( 4x - 4 ) = 0
<=> \(\orbr{\begin{cases}2x-6=0\\4x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
4) ( x2 - 2x )2 - 2( x - 1 )2 + 2 = 0
<=> ( x2 - 2x )2 - 2( x2 - 2x + 1 ) + 2 = 0
<=> ( x2 - 2x )2 - 2x2 + 4x - 2 + 2 = 0
<=> ( x2 - 2x )2 - 2( x2 - 2x ) = 0
<=> ( x2 - 2x )( x2 - 2x - 2 ) = 0
<=> \(\orbr{\begin{cases}x^2-2x=0\\x^2-2x-2=0\end{cases}}\)
+) x2 - 2x = 0 <=> x( x - 1 ) = 0 <=> \(\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
+) x2 - 2x - 2 = 0
<=> x2 - 2x + 1 - 3 = 0
<=> ( x2 - 2x + 1 ) = 3
<=> ( x - 1 )2 = ( ±√3 )2
<=> \(\orbr{\begin{cases}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{cases}}\)
10 tháng 3 2017
Thêm nữa câu a) Tính: M(x) + N(x)+ P(x)
B) Tính M(x) - N (x) - P(x)
ok rồi giúp mình với nha
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) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
6 tháng 8 2020
a, x=-505
b, x=35/8 hoac -37/8
nhung cau con lai thi tong tu
25 tháng 7 2019
a ) ( x + 1 )3 = 8
=> ( x + 1 )3 = 23
=> x + 1 = 2
=> x = 1
Vậy x = 1
b ) ( 2x + 3 )3 = 125
=> ( 2x + 3 )3 = 53
=> 2x + 3 = 5
2x = 5-3
2x = 2
x = 2:2 = 1
Vậy x = 1
c ) 2x + 1 hay 2x+1 ????
b. 1404 : [118 - (4x + 6)] = 27
118 - (4x + 6) = 52
4x + 6 = 66
4x = 60
x = 15
d) \(5x^2-3x=0\)
\(\Leftrightarrow x\left(5x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\5x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{5}\end{cases}}\)
e) \(3\left(x-1\right)+4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[3-4.\left(x-1\right)\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\3-4\left(x-1\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\4\left(x-1\right)=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x-1=\frac{3}{4}\Rightarrow x=\frac{7}{4}\end{cases}}\)
f) \(2\left(x-2\right)^2=\left(x-2\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2\left(x-2\right)-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x-2=\frac{1}{2}\Rightarrow x=\frac{5}{2}\end{cases}}\)
g) \(\left(x-2020\right)^4=\left(x-2020\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-2020\right)^2=0\\\left(x-2020\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=2019,x=2021\end{cases}}\)