(x-3)^x-(x-3)^x+2
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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 \(x=-\dfrac{11}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>9x^2+12x+4-9x^2+12x-4=5x+38
=>24x=5x+38
=>19x=38
=>x=2
e: =>x^3+1-2x=x^3-x
=>-2x+1=-x
=>-x=-1
=>x=1
f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1
=>12x-9=3x+1
=>9x=10
=>x=10/9
b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
=>-3x+3=3x-9
=>-6x=-12
=>x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
4 x 2 x 5 = ?
Cách 1: 4 x 2 x 5 = (4 x 2) x 5 = 8 x 5 = 40
Cách 2: 4 x 2 x 5 = 4 x (2 x 5) = 4 x 10 = 40
7 x 2 x 3 = ?
Cách 1: 7 x 2 x 3 = (7 x 2) x 3 = 14 x 3 = 42
Cách 2: 7 x 2 x 3 = 7 x (2 x 3) = 7 x 6 = 42
6 x 3 x 3 = ?
Cách 1: 6 x 3 x 3 = (6 x 3) x 3 = 18 x 3 = 54
Cách 2: 6 x 3 x 3 = 6 x (3 x 3) = 6 x 9 = 54
6 x 2 x 4 = ?
Cách 1: 6 x 2 x 4 = (6 x 2) x 4 = 12 x 4 = 48
Cách 2: 6 x 2 x 4 = 6 x (2 x 4) = 6 x 8 = 48
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
$(x-2)(x-5)=(x-3)(x-4)$
$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)
Vậy pt vô nghiệm.
2.
$(x-7)(x+7)+x^2-2=2(x^2+5)$
$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$
$\Leftrightarrow -51=10$ (vô lý)
Vậy pt vô nghiệm.
3.
$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$
$\Leftrightarrow 4x+10=-8$
$\Leftrightarrow 4x=-18$
$\Leftrightarrow x=-4,5$
4.
$(x+1)^2=(x+3)(x-2)$
$\Leftrightarrow x^2+2x+1=x^2+x-6$
$\Leftrightarrow x=-7$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A = 3 ( x − y ) 2 − 2 ( x + y ) 2 − ( x − y ) ( x + y ) 2 A = [ ( x − y ) − ( x + y ) ] 2 + 5 ( x − y ) 2 − 5 ( x + y ) 2 2 A = 4 y 2 + 5 [ ( x − y ) − ( x + y ) ] [ ( x − y ) + ( x + y ) ] 2 A = 4 y 2 + 5 [ − 2 y ] [ 2 x ] = 4 y 2 − 20 x y = 4 y ( y − 5 x ) A = 2 y ( y − 5 x )
(x-3)^x-(x-3)^x+2
= 0 + 2
= 2
bước 1 x-3x-x-3x2
-3x-3x2
bước 2 -3x2-3x
-3x2-3x
bước 3 x-3x-x-3x2
-3 (x2+x)
bước 4 -3 (x2+x)
-3x (x+1)
kết quả : -3x (x+1)