1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

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):}`

mk mới học lớp 6

\(\sqrt{2x}=4\)

\(\Leftrightarrow2x=4^2\) (ĐK: \(x\ge0\))

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=\dfrac{16}{2}\)

\(\Leftrightarrow x=8\left(tm\right)\)

Vậy \(x=8\)

    \(\sqrt{x}\) = 4
↔(\(\sqrt{x}\))\(^2\) = 4\(^2\)
↔ x = 16

Thay \(m=-2\) vào pt :

\(\Rightarrow\left(-2+1\right)x^2-3x+5=0\)

\(\Rightarrow-x^2-3x+5=0\)

\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-1\right)5=29>0\)

\(\Rightarrow\)Pt có 2 nghiệm phân biệt

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{29}}{-2}=\dfrac{-3+\sqrt{29}}{2}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{3-\sqrt{29}}{-2}=\dfrac{-3-\sqrt{29}}{2}\end{matrix}\right.\)

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

1/a/\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-6\end{cases}}}\)

Vậy ...................

b/ ĐKXĐ:\(x\ne2;x\ne5\)

.....\(\Rightarrow3x^2-15x-x^2+2x+3x=0\)

\(\Leftrightarrow2x^2-10x=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(nhận\right)\\x=5\left(loại\right)\end{cases}}}\)

Vậy ..............

`Answer:`

`1.`

a. \(\left(x+5\right)\left(2x+1\right)-x^2+25=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1-x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-5\end{cases}}}\)

b. \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\left(ĐKXĐ:x\ne2;x\ne5\right)\)

\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\)

\(\Leftrightarrow\frac{3x\left(x-5\right)-x\left(x-2\right)+3x}{\left(x-2\right)\left(x-5\right)}=0\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)+3x=0\)

\(\Leftrightarrow3x^2-15x-x^2+2x+3x=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\text{(Không thoả mãn)}\end{cases}}}\)

`2.`

\(ĐKXĐ:x\ne-m-2;x\ne m-2\)

Ta có: \(\frac{x+1}{x+2+m}=\frac{x+1}{x+2-m}\left(1\right)\)

a. Khi `m=-3` phương trình `(1)` sẽ trở thành: \(\frac{x+1}{x-1}=\frac{x+1}{x+5}\left(x\ne1;x\ne-5\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\frac{1}{x-1}=\frac{1}{x+5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=x+5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\-1=5\text{(Vô nghiệm)}\end{cases}}}\)

b. Để phương trình `(1)` nhận `x=3` làm nghiệm thì

\(\Leftrightarrow\hept{\begin{cases}\frac{3+1}{3+2-m}=\frac{3+1}{3+2-m}\\3\ne-m-2\\3\ne m-2\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{5+m}=\frac{4}{5-m}\\m\ne\pm5\end{cases}}\Leftrightarrow\hept{\begin{cases}5+m=5-m\\m\ne\pm5\end{cases}}\Leftrightarrow m=0\)