tìm x (x+2)(x^2-2x+4)-(x(x+3)(x-3)=26
tìm x 4.(x+1).(-x+2)+(2x-1).(2x+3)=-11 (2x+4).(3x+1).(x-2)-(-3x2 +1).(-2x+2/3)=-26/3
\(4\left(x+1\right)\left(-x+2\right)+\left(2x-1\right)\left(2x+3\right)=-11\)
\(\text{⇔}-4x^2+4x+8+4x^2+4x-3=-11\)
\(\text{⇔}8x+5=-11\)
\(\text{⇔}8x=-16\)
\(\text{⇔}x=-2\)
Vậy: \(x=-2\)
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\(\left(2x+4\right)\left(3x+1\right)\left(x-2\right)-\left(-3x^2+1\right)\left(-2x+\dfrac{2}{3}\right)=-\dfrac{26}{3}\)
\(\text{⇔}6x^3+2x^2-24x-8-6x^3-2x^2-2x+\dfrac{2}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x-\dfrac{22}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x=-\dfrac{4}{3}\)
\(\text{⇔}x=\dfrac{2}{39}\)
Tìm x, biết: (x+2)(x2-2x+4)-x(x-3)(x+3)=26
(x+2)(x2-2x+4)-x(x-3)(x+3)
=x3+8-x(x2-9)
=x3+8-x3+9x
=9x+8
Vậy 9x+8=26 => 9x=26-8 => 9x=18 => x=2
tìm x (2x+4).(3x+1).(x-2)-(-3x2 +1).(-2x+2/3)=-26/3
Bài 1: Tìm x:
1) (x-3)3 -( x-3)(x2+ 3x+9) +6( x+1)2+ 3x2 = -33
2) (X-3)( X2+ 3X+9) - X(X-2)( 2+X) = 1
3) (X+2)(X2 - 2X+4) – X(X-3)(X+3) = 26
a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)
\(\Leftrightarrow39x=-34\)
hay \(x=-\dfrac{34}{39}\)
b: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x=28\)
hay x=7
c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
\(\Leftrightarrow x=2\)
Tìm x: (x+2).(x^2-2x+4-x).(x+3).(x-3)=26
Giúp mik với. mik đg cần gấp
Tìm x: a) 2x(x-3)-x(3+2x)=26
b) (x+4)2-(x+1)(x-1)=16
c) (2x-1)2-4(x+7)(x-7)=0
d) x(x-5)-4x+20=0
mk làm lun nha
a, 2x^2-6x-3x-2x^2=26
-9x=26
x=-26/9
b,x^2+2.x.4+16-(x^2-1)=16
x^2+8x+16-x^2+1=16
8x=-1
x=-1/8
c,(2x)^2-2.2x.1+1-4(x^2-7^2)=0
4x^2-4x+1-4x^2+196=0
-4x=-197
x=197/4
d,x^2-5x-4x+20=0
-9x=-20
x=20/9
**** cho mk nha
Bài 19 Rút gọn
1) (x+2)^2+(3-x)^2
2) (4-x)^2 -(x-3)^2
3) (x-5)(x+5)-(x+5)^2
4) (x-3)^2-(x-4)(x+4)
5) (y^2 -6y+9)-(3-y)^2
6. (2x+3)² –(2x–3).(2x+3)
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
=-10x-50
4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16\)
=-6x+25
5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)
\(=y^2-6y+9-y^2+6y-9\)
=0
6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
=12x+18
Bài 19 rút gọn
1) (x+2)^2+(3-x)^2
2) (4-x)^2-(x-3)^2
3) (x-5)(x+5)-(x+5)^2
4)(x-3)^2-(x-4)(x+4)
5) (y^2-6y+9)-(3-y)^2
6) (2x+3)^2-(2x-3)(2x+3)
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=\left(-2x+7\right)\cdot1\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
\(=-10x-50\)
Bài 1: tìm x biết
a, (3x+4)^2 - (3x-1) (3x+1)=49
b, (x+2) (x^2x+4) -x (x+3) (x-3)=26