9 x 3 - 9 x + 1 x + 3 : x - 3 x 2 + 3 x - x 3 x + 9
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![](https://rs.olm.vn/images/avt/0.png?1311)
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.
$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$
$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$
$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$
$\Leftrgihtarrow 24x=22$
$\Leftrightarrow x=\frac{11}{12}$
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= (45 – 45) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0
b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 64 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0
= 0
c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)
= (36 – 36) : (3 x 5 x 7 x 9 x 11)
= 0 : (3 x 5 x 7 x 9 x 11)
= 0
d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7
= (27 – 27) : 9 x 1 x 3 x 5 x 7
= 0 : 9 x 1 x 3 x 5 x 7
=0
a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7
b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0
c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)
= 0 : (3 x 5 x 7 x 9 x 11)
d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7
= 0 : 9 x 1 x 3 x 5 x 7 Nếu đúng thì k cho mình nhé bạn!
![](https://rs.olm.vn/images/avt/0.png?1311)
Em ơi em đăng câu hỏi thì dùng công thức toán học để viết chữ viết thế này ai hiểu đúng được em?
![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
![](https://rs.olm.vn/images/avt/0.png?1311)
1) `x^2+4-2(x-1)=(x-2)^2`
`<=>x^2+4-2x+2=x^2-4x+4`
`<=>-2x+2=-4x`
`<=>2x=-2`
`<=>x=-1`
.
2) ĐKXĐ: `x \ne \pm 3`
`(x+3)/(x-3)-(x-1)/(x+3)=(x^2+4x+6)/(x^2-9)`
`<=>(x+3)^2-(x-1)(x-3)=x^2+4x+6`
`<=>x^2+6x+9-x^2+4x-3=x^2+4x+6`
`<=>10x+6=x^2+4x+6`
`<=>x^2-6x=0`
`<=>x(x-6)=0`
`<=>x=0;x=6`
.
3) ĐKXĐ: `x \ne \pm 3`
`(3x-3)/(x^2-9) -1/(x-3 )= (x+1)/(x+3)`
`<=>(3x-3)-(x+3)=(x+1)(x-3)`
`<=> 2x-6=x^2-2x-3`
`<=>x^2-4x+3=0`
`<=>x^2-x-3x+3=0`
`<=>x(x-1)-3(x-1)=0`
`<=>(x-3)(x-1)=0`
`<=> x=3;x=1`
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
\(\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) ĐKXĐ: \(x\notin\left\{-2;2\right\}\)
Ta có: \(\dfrac{x-1}{x+2}-\dfrac{9}{x^2-4}=\dfrac{-3}{x-2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{9}{\left(x-2\right)\left(x+2\right)}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2-3x+2-9=-3x-6\)
\(\Leftrightarrow x^2-3x-7+3x+6=0\)
\(\Leftrightarrow x^2-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1;-1}
2)
Sửa đề: \(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)
ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)
\(\Leftrightarrow\dfrac{3x-3}{\left(x-3\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
Suy ra: \(3x-3-x-3=x^2-3x+x-3\)
\(\Leftrightarrow x^2-2x-3=2x-6\)
\(\Leftrightarrow x^2-2x-3-2x+6=0\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
Vậy: S={1}
`1)(x-1)/(x+2)-9/(x^2-4)=-3/(x-2)(x ne 2)`
`<=>x^2-3x+2-9=-3x-6`
`<=>x^2-1=0`
`<=>x=+-1`