24x²-12x-24x²+18x=30

6x=30

x=5

Ta có \(3x\left(8x-4\right)-6x\left(4x-3\right)=30\)

\(\Rightarrow24x^2-12x-24x^2+18x=30\)

\(\Rightarrow6x=30\)

\(\Rightarrow x=\frac{30}{6}=5\)

Vậy phương trình có tập nghiệm S={5}

\(\Leftrightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x-5\right)\left(x+5\right)}+\dfrac{7}{6\left(x+5\right)}=0\)

\(\Leftrightarrow3\cdot3\left(x+5\right)-15\cdot6+7\cdot2\cdot\left(x-5\right)=0\)

=>9x+45-90+14x-70=0

=>23x-115=0

hay x=5(loại)

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

`3/(4x-20)+15/(50-2x^2)+7/(6x+30)=0(x ne +-5)`

`pt<=>3/(4(x-5))+15/(2(5-x)(5+x))+7/(6(x+5))=0`

`<=>3/(4(x-5))+7/(6(x+5))-15/(2(x-5)(x+5))=0`

`<=>9(x+5)+14(x-5)-90=0`

`<=>9x+45+14x-70-90=0`

`<=>23x=115`

`<=>x=5(ktm)`

Vậy PTVN

a: =(x-2)(3x-2)

c)x²-2x+2

=x²-2x+1+1

=(x-1)²+1

=(x-1)²-i²

^{=[(x-1)-i][(x-1)+i}

=(x-1-i)(x-1+i)

b)9x²-6x+5

=9x²-3x-3x+5

=3x(3x-5)-5(3x-5)

=(3x-1)²

c)30-20x+4x²

^{chịu ,khó was ! k lm dc !!!!!!!!!!!!!!!!!!}

a) \(4x^2-4x-15=0\)

<=>  \(\left(2x-5\right)\left(2x+3\right)=0\)

<=>  \(\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\)<=>  \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)

Vậy....

b)  \(x^3-6x^2-x+30=0\)

<=>  \(\left(x-5\right)\left(x-3\right)\left(x+2\right)=0\)

tự giải tiếp

ĐKXĐ: \(\left\{{}\begin{matrix}4x-20\ne0\\50-2x^2\ne0\\6x+30\ne0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}4x-20\ne0\\x^2-25\ne0\\6x+30\ne0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x-5\ne0\\x+5\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)

=> \(x\ne\pm5\)

Ta có : \(\frac{3}{4x-20}+\frac{15}{50-2x^2}+\frac{7}{6x+30}=0\)

=> \(\frac{3}{4\left(x-5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}+\frac{7}{6\left(x+5\right)}=0\)

=> \(\frac{9\left(x+5\right)}{12\left(x^2-25\right)}-\frac{90}{12\left(x^2-25\right)}+\frac{14\left(x-5\right)}{12\left(x^2-25\right)}=0\)

=> \(9\left(x+5\right)-90+14\left(x-5\right)=0\)

=> \(9x+45-90+14x-70=0\)

=> \(23x=115\)

=> \(x=5\) ( KTM )

Vậy phương trình vô nghiệm .