Giải phương trình:[(x^2+x-5)/x]+[3x/(x^2+x-5)]+4=0

Đặt (x^2+x-5)/x = a ta có phương trình :

a + 3/a + 4 = 0 (a#0) <=> a^2 + 4a + 3 = 0 <=> a=-3 hoặc a=-1

sau đó thế vào giải là ra nha

( điều kiện xác định thì bạn tự làm nha )

\(\frac{x^2+x-5}{x}\)+\(\frac{3x}{x^2+x-5}\)+4=0

ĐKXĐ: x\(\ne\)0 và x \(\ne\)\(\frac{\sqrt{21}-1}{2}\)và x\(\ne\)-\(\frac{\sqrt{21}-1}{2}\)

Ta có:

\(\frac{x^2+x-5}{x}\)+ \(\frac{3x}{x^2+x-5}\)+4=0

<=> \(\frac{\left(x^2+x-5\right)^2}{x\left(x^2+x-5\right)}\)+\(\frac{3x^2}{x\left(x^2+x-5\right)}\)+\(\frac{4x\left(x^2+x-5\right)}{x\left(x^2+x-5\right)}\)=0

<=> \(\frac{x^4+6x^3-2x^2-30x+25}{x\left(x^2+x-5\right)}\)=0

=> x\(^4\)+6x\(^3\)-2x\(^2\)-30x+25=0

<=> x⁴+4x³+2x³+8x²-10x²-40x+10x+40=15

<=> (x+4)(x³+2x²-10x+10)=15

Với x+4 =1 thì x=-3 => x³+2x²-10x+10= 31 (loại)

Với x+4=-1 thì x=-5 => x³+2x²-10x+10=-15 (thỏa mãn)

.........

Có 6 trường hợp , ta có: x=-5 hoặc x= 1

Vậy x=-5 hoặc x=1

a.

\(x^2-x-\left(5x-5\right)=0\)

\(\Leftrightarrow x\left(x-1\right)-5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

Câu b hoàn toàn tương tự

Bài `1:`

`h)(3/4x-1)(5/3x+2)=0`

`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`

______________

Bài `2:`

`b)3x-15=2x(x-5)`

`<=>3(x-5)-2x(x-5)=0`

`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`

`d)x(x+6)-7x-42=0`

`<=>x(x+6)-7(x+6)=0`

`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`

`f)x^3-2x^2-(x-2)=0`

`<=>x^2(x-2)-(x-2)=0`

`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`

`h)(3x-1)(6x+1)=(x+7)(3x-1)`

`<=>18x^2+3x-6x-1=3x^2-x+21x-7`

`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`

`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`

`j)(2x-5)^2-(x+2)^2=0`

`<=>(2x-5-x-2)(2x-5+x+2)=0`

`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`

`w)x^2-x-12=0`

`<=>x^2-4x+3x-12=0`

`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`

`m)(1-x)(5x+3)=(3x-7)(x-1)`

`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`

`<=>(1-x)(5x+3+3x-7)=0`

`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`

`p)(2x-1)^2-4=0`

`<=>(2x-1-2)(2x-1+2)=0`

`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`

`r)(2x-1)^2=49`

`<=>(2x-1-7)(2x-1+7)=0`

`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`

`t)(5x-3)^2-(4x-7)^2=0`

`<=>(5x-3-4x+7)(5x-3+4x-7)=0`

`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`

`u)x^2-10x+16=0`

`<=>x^2-8x-2x+16=0`

`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`

\(\dfrac{x+2}{x-5}-3< 0\)

\(\Leftrightarrow\dfrac{x+2-3\left(x-5\right)}{x-5}< 0\)

\(\Leftrightarrow x+2-3x+15< 0\)

\(\Leftrightarrow-2x+17< 0\)

\(\Leftrightarrow-2x< -17\)

\(\Leftrightarrow x>\dfrac{17}{2}\)

\(\left(x-1\right)\left(4-x\right)\ge x\left(x-3\right)-2x^2\)

\(\Leftrightarrow4x-x^2-4+x-x^2+3x+2x^2\ge0\)

\(\Leftrightarrow8x-4\ge0\)

\(\Leftrightarrow4\left(2x-1\right)\ge0\)

\(\Leftrightarrow2x-1\ge0\)

\(\Leftrightarrow2x\ge1\)

\(\Leftrightarrow x\ge\dfrac{1}{2}\)

\(\left(2x+5\right)\left(x-4\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(\left(2x+5\right)+3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+5+3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\2x+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\2x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{8}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(\left(2x+5\right)\left(x-4\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left[\left(2x+5\right)+3\right]=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\2x+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

CHỊU!@@@@@@@@@@@@

a)

\(x^2-4\sqrt{15}x+19=0\\ < =>x^2-4\sqrt{15}x+60-41=0\\ < =>\left(x-2\sqrt{15}\right)^2-41=0\\ < =>\left(x-2\sqrt{15}-\sqrt{41}\right)\left(x-2\sqrt{15}+\sqrt{41}\right)=0\\ < =>\left[{}\begin{matrix}x-2\sqrt{15}-\sqrt{41}=0\\x-2\sqrt{15}+\sqrt{41}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=2\sqrt{15}+\sqrt{41}\\x=2\sqrt{15}-\sqrt{41}\end{matrix}\right.\)

b)

\(4x^2+4\sqrt{5}x+5=0\\ < =>\left(2x+\sqrt{5}\right)^2=0\\ < =>2x+\sqrt{5}=0\\ < =>2x=-\sqrt{5}\\ < =>-\dfrac{\sqrt{5}}{2}\)

a: Δ=(4căn 15)^2-4*1*19=164>0

Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{5}-2\sqrt{41}}{2}=2\sqrt{5}-\sqrt{41}\\x_2=2\sqrt{5}+\sqrt{41}\end{matrix}\right.\)

b: \(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot\sqrt{5}+5=0\)

=>(2x+căn 5)^2=0

=>2x+căn 5=0

=>x=-1/2*căn 5