Tìm x ϵ N,biết
a)x2 - 5 = 11
b)4x3 + 15 = 19
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KK
12 tháng 9 2021
2.
a. 3x(12x - 4) - 9x(4x - 3) = 30
<=> 36x2 - 12x - 36x2 + 27x = 30
<=> 36x2 - 36x2 - 12x + 27x = 30
<=> 15x = 30
<=> x = 2
b. x(5 - 2x) + 2x(x - 1) = 15
<=> 5x - 2x2 + 2x2 - 2x = 15
<=> -2x2 + 2x2 + 5x - 2x = 15
<=> 3x = 15
<=> x = 5
PT
Phạm Trần Hoàng Anh
CTVHS
12 tháng 9 2021
a) x2 ( 5x3 - x - 1212)= 5x5-x3-1212x
b) ( 3xy - x2 + y ) 2323x2y= 6969x3y2- 2323x4y+ 2323x2y2
c) x2 ( 4x3 - 5xy + 2x ) ( -1212 xy )=(4x5-5x3y+2x3).(-1212xy)
= -4848x6y +6060x4y2-2424x4y
2/ Tìm x, biết
a) 3x( 12x - 4 ) - 9x (4x - 3 ) = 30
=> 36x2-12x-36x2+27x=30
=> -12x +27x=30
=> 15x = 30
=>x =2
b ) x( 5 - 2x ) + 2x ( x - 1 )= 15
=> 5x-2x2+2x2-2x=15
=> 3x=15
=>x=5
T
1
28 tháng 9 2021
a) \(\left(x-2\right)\left(y+1\right)=14\)
Do \(x,y\in N\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=1\\y+1=14\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=14\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=2\\y+1=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=7\\y+1=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\left(tm\right)\\y=13\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=16\left(tm\right)\\y=0\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\left(tm\right)\\y=6\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=9\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
22 tháng 10 2023
a: 17-2x=9
=>2x=17-9=8
=>x=8/2=4
b: \(145-135\left(x-2\right)^2=10\)
=>\(135\cdot\left(x-2\right)^2=135\)
=>\(\left(x-2\right)^2=1\)
=>\(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c: \(x\inƯ\left(36\right)\)
=>\(x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)
mà x>12
nên \(x\in\left\{18;36\right\}\)
d: \(x-1\in B\left(9\right)\)
=>\(x-1\in\left\{0;9;18;27;36;45;54;...\right\}\)
=>\(x\in\left\{1;10;19;28;37;46;55;...\right\}\)
mà 25<x<50
nên \(x\in\left\{28;37;46\right\}\)
21 tháng 7 2021
Bài 10:
a) (x+2)2 -x(x+3) + 5x = -20
=> x2 + 4x + 4 - x2 - 3x + 5x = -20
=> 6x = -20 + (-4)
=> 6x = -24
=> x = -4
b) 5x3-10x2+5x=0
=>5x(x2-2x+1)=0
=>5x(x-1)2 =0
=> 5x=0 hoặc (x-1)2=0
=>x=0 hoặc x=1
c) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
=> (x2 - 1)[(x2 - 1)2 - (x4 + x2 + 1)] = 0
<=> (x2 - 1)(x4 - 2x2 + 1 - x4 - x2 - 1) = 0
<=> (x2 - 1)(-3x2) = 0
<=> (x2 - 1)=0 hoặc (-3x2) =0
<=> x2=1 hoặc x2=0
<=> x=−1;1 hoặc x=0
d)
(x+1)3−(x−1)3−6(x−1)2=-19
⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0
⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0
⇔12x+13=0⇔12x+13=0
⇔12x=-13
⇔x=-23/12
Học tốt nhé:333
20 tháng 3 2023
1:
a: \(=\dfrac{-4}{7}+\dfrac{4}{7}+\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=-\dfrac{1247}{1190}\)
b:
Sửa đề: \(\dfrac{-5}{13}+\dfrac{4}{19}+\dfrac{-8}{13}+\dfrac{15}{19}+\dfrac{45}{6}\)
\(=\dfrac{-5}{13}-\dfrac{8}{13}+\dfrac{4}{19}+\dfrac{15}{19}+\dfrac{45}{6}=\dfrac{9}{2}\)
3 tháng 10 2021
a: Ta có: \(A=2x^2+12x+11\)
\(=2\left(x^2+6x+\dfrac{11}{2}\right)\)
\(=2\left(x^2+6x+9-\dfrac{7}{2}\right)\)
\(=2\left(x+3\right)^2-7\ge-7\forall x\)
Dấu '=' xảy ra khi x=-3
3 tháng 10 2021
\(A=2\left(x^2+6x+36\right)-61=2\left(x+6\right)^2-61\ge-61\\ A_{min}=-61\Leftrightarrow x=-6\\ B=-\left(x^2-18x+81\right)+100=-\left(x-9\right)^2+100\le100\\ B_{max}=100\Leftrightarrow x=9\)
NH
2
K
10 tháng 12 2021
\(35-5\left(x-1\right)=10\\ \Leftrightarrow35-5x+5=10\\ \Rightarrow40-5x=10\)
\(\Rightarrow-5x=10-40\\ \Rightarrow-5x=-30\\ \Rightarrow x=\dfrac{-30}{-5}=6\)
c)
\(24\left(x-16\right)=12^2\)
\(\Rightarrow24x-384=144\\ \Rightarrow24x=144+384\\ \Rightarrow24x=528\\ \Rightarrow x=\dfrac{528}{24}=22\)
d)
\(\left(x^2-10\right)\div5=3\\ \Rightarrow\left(x^2-10\right)=3\times5\\ \Rightarrow x^2-10=15\)
\(\Rightarrow x^2=15+10\\ \Rightarrow x^2=25\\ \Rightarrow x^2=5^2\Rightarrow x=5\)
22 tháng 12 2021
a) \(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\Rightarrow\left(x-1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a ) \(x^2-5=11\)
\(\Leftrightarrow x^2=11+5\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm\sqrt{16}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-4\end{array}\right.\)
Vậy \(x\in\left\{4;-4\right\}\)
b ) \(4x^3+15=19\)
\(\Leftrightarrow4x^3=19-15\)
\(\Leftrightarrow4x^3=4\)
\(\Leftrightarrow x^3=4:4\)
\(\Leftrightarrow x^3=1\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
a. x^2=16
x^2=4^2
x=4
b.4x^3=4
x^3=1
x=1