tìm x,y
a) |x-3,5|+|4,5-x|=0
b) |x2-2x|=x
Bài 1: Tính nhanh
a) 15.64 + 25.100 + 36.15 + 60.1000
b) 37,5. 6,5 – 7,5 .3,4 -6,6 .7,5 +3,5. 37,5
Bài 2: Tìm x
a) x.(x-2) + x - 2 = 0
b) 5x.(x-3) - x+3 =0
Bài 3 Phân tích các đa thức sau thành nhân tử
a) x3 - 2x + x
b) x2 + 2x +1
c) x2 -16
d) (2x-1)2 –(x+3)2
Bài 3:
b: \(x^2+2x+1=\left(x+1\right)^2\)
c: \(x^2-16=\left(x-4\right)\left(x+4\right)\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2\)
\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)
4/ Tìm y
A/ 25,5 + y - 12,5 = 4 x 7
b. 76,22 - y - 25,7 = 30 + 5,52
c. 4,5 - y +1,2 = 3,5
a)\(25,5+y-12,5=4.7\)
⇔\(13+y=28\)
⇔\(y=15\)
b)\(76,22-y-25,7=30+5,52\)
⇔\(50,52-y=35,52\)
⇔\(y=15\)
c)\(4,5-y+1,2=3,5\)
⇔\(5,7-y=3,5\)
⇔\(y=2,2\)
\(a,\Rightarrow10+y=28\\ \Rightarrow y=18\\ b,\Rightarrow50,52-y=35,52\\ \Rightarrow y=15\\ c,\Rightarrow5,7-y=3,5\\ \Rightarrow y=2,2\)
a) \(25,5+y-12,5=4\times7\)
\(\Rightarrow25,5+y-12,5=28\)
\(\Rightarrow y-12,5=28-25,5\)
\(\Rightarrow y-12,5=2,5\)
\(\Rightarrow y=2,5+12,5\)
\(\Rightarrow y=15\)
b) \(76,22-y-25,7=30+5,52\)
\(\Rightarrow76,22-y-25,7=35,52\)
\(\Rightarrow y-25,7=76,22-35,52\)
\(\Rightarrow y-25,7=40,7\)
\(\Rightarrow y=40,7+25,7\)
\(\Rightarrow y=66,4\)
c) \(4,5-y+1,2=3,5\)
\(\Rightarrow4,5-y=3,5-1,2\)
\(\Rightarrow4,5-y=2,3\)
\(\Rightarrow y=4,5-2,3\)
\(\Rightarrow y=2,2\)
Tìm x:
x(2x-1)(x+5)-(2x^2+1)(x+4,5)=3,5
Tìm x
a) x2(x+1)+x+1=0
b) x2-x=-2x2+2x
c) 2x2(x-1)+x2=x
d) (x-2)(x2+4)=x2-2x
a) Ta có: \(x^2\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
b) Ta có: \(x^2-x=-2x^2+2x\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
c) Ta có: \(2x^2\left(x-1\right)+x^2=x\)
\(\Leftrightarrow2x^2\left(x-1\right)+x^2-x=0\)
\(\Leftrightarrow2x^2\left(x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\cdot\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
d) Ta có: \(\left(x-2\right)\left(x^2+4\right)=x^2-2x\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4\right)-x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x+4\right)=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Tìm x,biết:
a)2x.(x+4)-(x-1).(2x+3)=0
b)x2-2x-3=0
a) \(2x\left(x+4\right)-\left(x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x^2+8x-2x^2-x+3=0\)
\(\Leftrightarrow7x=-3\Leftrightarrow x=-\dfrac{3}{7}\)
b) \(x^2-2x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
\(a,\Leftrightarrow2x^2+8x-2x^2-x+3=0\\ \Leftrightarrow7x=-3\\ \Leftrightarrow x=-\dfrac{3}{7}\\ b,x^2-2x-3=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
a: Ta có: \(2x\left(x+4\right)-\left(x-1\right)\cdot\left(2x+3\right)=0\)
\(\Leftrightarrow2x^2+8x-2x^2-3x+2x+3=0\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
b: ta có: \(x^2-2x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Tìm x:
a)2x3-18x=0
b)(3x-2).(2x+1)-6x.(x+2)=11
c)(x-1)3-(x+2).(x2-2x+4)=3.(1-x2)
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
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
c. (x - 1)3 - (x + 2)(x2 - 2x + 4) = 3(1 - x2)
<=> (x3 - 3x2 + 3x - 1) - (x3 - 2x2 + 4x + 2x2 - 4x + 8) = 3 - 3x2
<=> x3 - 3x2 + 3x - 1 - x3 + 2x2 - 4x - 2x2 + 4x - 8 = 3 - 3x2
<=> x3 - x3 - 3x2 + 2x2 - 2x2 + 3x2 + 3x - 4x + 4x = 3 + 1 + 8
<=> 3x = 12
<=> x = 4
Tìm x
a, 3/4x*(x2-9)=0
b, x3-16x=0
c, (x-1)(x+2)-x-2=0
d, 3x3-27x=0
e, x2(x+1)+2x(x+1)=0
f, x(2x-3)-2(3-2x)=0
c: =>(x-1)(x+1)=0
hay \(x\in\left\{1;-1\right\}\)
a,
\(=\dfrac{3}{4x}.\left(x-3\right)\left(x+3\right)\)=0
\(\left\{{}\begin{matrix}\dfrac{3}{4x}=0\\x-3=0\\x+3=0\end{matrix}\right.\)
=>\(x=\left\{3,-3\right\}\)
b,
\(x^3-16x=0\\x\left(x^2-16\right)\\ x\left(x-4\right)\left(x+4\right)\)
\(\left\{{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
=>\(x=\left\{-4,0,4\right\}\)
d,
\(3x^3-27x=0\\ 3x\left(x^2-9\right)=0\\ 3x\left(x-3\right)\left(x+3\right)=0\)
\(\left\{{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\)
=>\(x=\left\{-3,0,3\right\}\)
e,
\(x^2+\left(x+1\right)+2x\left(x+1\right)=0\\ x\left(x+1\right)\left(x+2\right)=0\)
\(\left\{{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)
=>\(x=\left\{-2,-1,0\right\}\)
f,
\(x\left(2x-3\right)-2\left(3-2x\right)=0\\ \left(2x-3\right)\left(x+2\right)=0\)
\(\left\{{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
tìm x, biết:
a)(x-3)2 - 4 = 0
b) x2 - 2x = 24
a. (x - 3)2 - 4 = 0
<=> (x - 3)2 - 22 = 0
<=> (x - 3 + 2)(x - 3 - 2) = 0
<=> (x - 1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
b. x2 - 2x = 24
<=> x2 - 2x - 24 = 0
<=> x2 - 6x + 4x - 24 = 0
<=> x(x - 6) + 4(x - 6) = 0
<=> (x + 4)(x - 6) = 0
<=> \(\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=4\)
TH1:\(x-3=2\text{⇒}x=5\)
TH2:\(x-3=-2\text{⇒}x=1\)
\(a\left(x-3\right)^2-4=0\)
\(\Rightarrow\left(x-3\right)^2-2^2=0\)
\(\Rightarrow\left(x-3-2\right).\left(x-3+2\right)=0\)
\(\Rightarrow\left(x-5\right).\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{1;5\right\}\)
\(b,x^2-2x=24\)
\(\Rightarrow x^2-2x-24=0\)
\(\Rightarrow x^2+4x-6x-24=0\)
\(\Rightarrow x.\left(x+4\right)-6.\left(x+4\right)\)
\(\Rightarrow\left(x-6\right).\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{-4;6\right\}\)
Hoctot
Tìm x:
a) x(x-2)-x+2=0
b) (3-2x)2-(x-1)2=0
c) 81x4-x2=0
d) x3+x2+27x+27=0