Tìm x:
2/3(x+1/2)+1/3(x+1 1phan2)=1x=-2;x=0 và x=2 có phải là các nghiệm của đa thức x^3-4x hay ko ?vì sao
trong các số cho ở bảng sau số nào là nghiệm của đa thức (ở cùng hàng) ?
a)p(x) =2x+1phan 2 | 1phan4 | 1phan2
| -1phan4 |
b)q(x)=x^2-2x-3 | 3 | 1
| -1 |
từ kết quả trên hãy trả lời câu hỏi :có thể tìm nghiệm của đa thức bằng cách nào?
(x-2phan5) nhan 1phan2 =1
\(\left(x-\frac{2}{5}\right)×\frac{1}{2}=1=>\left(x-\frac{2}{5}\right)=1:\frac{1}{2}=>\left(x-\frac{2}{5}\right)=2=>x=2+\frac{2}{5}=>x=\frac{12}{5}\)
vậy \(x=\frac{12}{5}\)
Bài làm :
Ta có :
\(\left(x-\frac{2}{5}\right)\times\frac{1}{2}=1\)
\(x-\frac{2}{5}=1\div\frac{1}{2}\)
\(x-\frac{2}{5}=2\)
\(x=2+\frac{2}{5}=2,4\)
Vậy x=2,4
= (x - 2/5) * 1/2 =1
=x- 2/5=1:1/2
=x -2/5= 2
=x = 2 +2/5
=x = 12/5
. Tìm x biết rằng:
a)(x + 1)3 – (x + 2)(x – 1)2 – 3(x – 3)(x + 3) = 5
b)(x + 1)3 + (x – 1)3 = (x + 2)3 + (x – 2)3
c) (x + 1)3 - (x - 1)3 - 6(x - 1)2 = -10
a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)
\(\Leftrightarrow6x=-3\)
hay \(x=-\dfrac{1}{2}\)
b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)
\(\Leftrightarrow2x^3+6x=2x^3+24x\)
\(\Leftrightarrow x=0\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)
\(\Leftrightarrow12x=-11\)
hay \(x=-\dfrac{11}{12}\)
Tìm x Toán đại 8 Hằng đẳng thức đáng nhớ?
Tìm x:
1. (x-1)^3+3.(x-3)^2-(x+2).(x^2-2x+4) = (x+2)^3-(x-3).(x^2+9)-6x^2+5
2.(3+2x)^3-(6x-1).(6x+1) = (2x-1)^3+(x+4)^2-x^3+(x+1).(x^2+x+1)
1. (x - 1)^3 + 3.(x - 3)^2 - (x + 2).(x^2 - 2x + 4) = (x + 2)^3 - (x - 3).(x^2 + 9) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3(x^2 - 6x + 9) - (x^3 + 2^3)
= x^3 + 6x^2 + 12x + 8 - (x^3 - 3x^2 + 9x -27) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3x^2 - 18x + 27 - x^3 - 8
= x^3 + 6x^2 + 12x + 8 - x^3 + 3x^2 - 9x + 27 - 6x^2 + 5
<=> 3x - 18x -12x - 3x^2 + 9x = 27 + 5 + 8 + 8 + 1 - 27
<=> - 3x^2 - 18x - 22 = 0
<=> 3x^2 + 18x + 22 = 0
Nửa chu vi mảnh đất là:
120 : 2 = 60 (m)
Chiều dài hơn chiều rộng là:
5 + 5 = 10 (m)
Chiều rộng là:
( 60 - 10 ) : 2 = 25 (m)
Chiều dài là:
25 + 10 = 35 (m)
Diện tích là:
25 35 = 875 ( )
tìm x
b) (x + 3)2 - (x - 4)( x + 8) = 1;
c) 3(x + 2)2 + (2x - 1)2 - 7(x + 3)(x - 3) = 36;
d)(x - 3)(x2 + 3x + 9) + x(x + 2)(2 - x) = 1;
e) (x + 1)3 - (x - 1)3 - 6(x - 1)2 = -19.
hihi
b)(x+3)2-(x-4)(x+8)=1
\(\Rightarrow\)x2+6x+9-(x2+8x-4x-32)=1
⇒x2+6x+9-x2-8x+4x+32=1
⇒2x+41=1
\(\Rightarrow\)2x+41-1=0
\(\Rightarrow\)2x+40=0
⇒2x=-40
\(\Rightarrow\)x=\(\dfrac{-40}{2}\)
⇒x=-20
Tìm x :
a) x (3x + 1) + (x -1)2 - (2x + 1)(2x -1) = 0
b) (x + 1)3 + (2 - x)3 - 9(x - 3)(x+3) = 0
c) (x - 1)3 - (x + 3)(x2 - 3x + 9) + 3x2 = 25
d) (x + 2)3 - ( x +1)(x2 - x + 1) - 6(x - 1)2 = 23
e) (x + 3)(x2 - 3x + 9) - x(x - 2)(x+2) + 11 = 0
f) x(x - 3) - x + 3 = 0
Lời giải:
a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$
$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$
$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$
$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$
$\Leftrightarrow -x+2=0$
$\Leftrightarrow x=2$
b.
$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$
$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$
$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$
$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$
$\Leftrightarrow -x+10=0\Leftrightarrow x=10$
c.
$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$
$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$
$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$
$\Leftrightarrow 3x-28=25$
$\Leftrightarrow x=\frac{53}{3}$
d.
$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$
$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$
$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$
$\Leftrgihtarrow 24x=22$
$\Leftrightarrow x=\frac{11}{12}$
e.
$(x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0$
$\Leftrightarrow x^3+3^3-x(x^2-4)+11=0$
$\Leftrightarrow x^3+27-x^3+4x+11=0$
$\Leftrightarrow (x^3-x^3)+4x+(27+11)=0$
$\Leftrightarrow 4x+38=0$
$\Leftrightarrow x=\frac{-19}{2}$
f.
$x(x-3)-x+3=0$
$\Leftrightarrow x(x-3)-(x-3)=0$
$\Leftrightarrow (x-3)(x-1)=0$
$\Leftrightarrow x-3=0$ hoặc $x-1=0$
$\Leftrightarrow x=3$ hoặc $x=1$
Bài 1:Tìm x,biết:
a,(x-2)(x+2)-(x-3)\(^2\)=9
b,(x-1)(x\(^2\)+1)-(x+1)(x\(^2\)-x+1)=x(2-x)
c,(x-3)(x\(^2\)+3x+9)+x(x+2)(2-x)=1
d,(x+1)\(^3\)-(x-1)\(^{^{ }3}\)-6(x-1)\(^2\)=-19
Bài 2:Viết về dạng bình phương hoặc dạng tích:
a,\(\dfrac{1}{27}\)x\(^3\)+x\(^2\)+9x+27
b,8u\(^3\)-60u\(^2\)v+150uv\(^2\)-125v\(^3\)
c,x^3+3x^2+3x+1+3(x^2+2x+1)y+3xy^2+3y^3+y^3
a. (x - 2)(x + 2) - (x - 3)2 = 9
<=> x2 - 22 - (x - 3)2 = 32
<=> x - 2 - (x - 3) = 3
<=> x - 2 - x + 3 = 3
<=> x - x = 3 - 3 + 2
<=> 0 = 2 (Vô lí)
Vậy nghiệm của PT là S = \(\varnothing\)
b: Ta có: \(\left(x-1\right)\left(x^2+1\right)-\left(x+1\right)\left(x^2-x+1\right)=x\left(2-x\right)\)
\(\Leftrightarrow x^3+x-x^2-1-x^3-1=2x-x^2\)
\(\Leftrightarrow-x^2+x-2-2x+x^2=0\)
\(\Leftrightarrow-x=2\)
hay x=-2
1.tìm x,y biết
a, x.(y-3)≥0
b, (2.x-1).(y-1)≤0
c,(x-1).(2.k+1)≥0
2. tìm x,y ϵ Z biết
a, x(x+3)=0
b,(x-2).(5-x)=0
c,(x-1).(x^2+1)=0
d, x.y+3.x-7.y=21
e,x.y+3.x-2y=11
Bài 2:
a: =>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
1) 3(x-2) + 4(x-1) = 25 2) (5x-3)(x-2) = (x-1)(x-2) 3) (x-2)² = 4(x-1)²
\(3\left(x-2\right)+4\left(x-1\right)=25\)
\(\Leftrightarrow3x-6+4x-4=25\)
\(\Leftrightarrow7x=35\)
\(\Leftrightarrow x=5\)
\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5x-3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{2}\end{matrix}\right.\)
\(\left(x-2\right)^2=4\left(x-1\right)^2\)
\(\Leftrightarrow\left(x-2\right)^2-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[\left(x-2\right)-2\left(x-1\right)\right]\left[\left(x-2\right)+2\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2-2x+2\right)\left(x-2+2x-2\right)=0\)
\(\Leftrightarrow\left(-x\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)