a: |x-1|+|x-2|=|2x-3|

=>|x-1|+|x-2|-|2x-3|=0(1)

TH1: x<1

=>x-1<0; 2x-3<0; x-2<0

(1) sẽ trở thành: 1-x+2-x-(3-2x)=0

=>3-2x-3+2x=0

=>0x=0(luôn đúng)

TH2: 1<=x<3/2

=>x-1>=0; 2x-3<0; x-2<0

(1) sẽ trở thành: x-1+2-x-(3-2x)=0

=>1-3+2x=0

=>2x-2=0

=>x=1(nhận)

TH3: 3/2<=x<2

=>x-1>0; 2x-3>=0; x-2<0

(1) sẽ trở thành: x-1+2-x-(2x-3)=0

=>1-2x+3=0

=>-2x+4=0

=>-2x=-4

=>x=2(loại)

TH4: x>=2

=>x-1>0; 2x-3>0; x-2>=0

(1) sẽ trở thành: x-1+x-2-(2x-3)=0

=>2x-3-2x+3=0

=>0x=0(luôn đúng)

Vậy: x<=1 hoặc x>=2

b: ĐKXĐ: x∉{2;3;5}

\(\frac{1}{x-2}+\frac{2}{x-3}-\frac{3}{x-5}=\frac{1}{x^2-5x+6}\)

=>\(\frac{1}{x-2}+\frac{2}{x-3}-\frac{3}{x-5}=\frac{1}{\left(x-2\right)\left(x-3\right)}=\frac{1}{x-3}-\frac{1}{x-2}\)

=>\(\frac{2}{x-2}+\frac{1}{x-3}-\frac{3}{x-5}=0\)

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

=>2(x-3)(x-5)+(x-2)(x-5)-3(x-2)(x-3)=0

=>\(2\left(x^2-8x+15\right)+x^2-7x+10-3\left(x^2-5x+6\right)=0\)

=>\(2x^2-16x+30+x^2-7x+10-3x^2+15x-18=0\)

=>-8x+22=0

=>-8x=-22

=>x=11/4(nhận)

1: 5(2-3x)(x-2)=3(1-3x)

=>5(3x-2)(x-2)=3(3x-1)

=>\(5\cdot\left(3x^2-6x-2x+4\right)=9x-3\)

=>\(15x^2-40x+20-9x+3=0\)

=>\(15x^2-49x+23=0\)

\(\Delta=49^2-4\cdot15\cdot23=1021>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{49-\sqrt{1021}}{2\cdot15}=\frac{49-\sqrt{1021}}{30}\\ x=\frac{49+\sqrt{1021}}{2\cdot15}=\frac{49+\sqrt{1021}}{30}\end{array}\right.\)

2: \(4x^2+4x+1=0\)

=>\(\left(2x+1\right)^2=0\)

=>2x+1=0

=>2x=-1

=>\(x=-\frac12\)

3: \(4x^2-9=0\)

=>\(4x^2=9\)

=>\(x^2=\frac94\)

=>\(\left[\begin{array}{l}x=\frac32\\ x=-\frac32\end{array}\right.\)

4: \(5x^2-10x=0\)

=>5x(x-2)=0

=>x(x-2)=0

=>\(\left[\begin{array}{l}x=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)

5: \(x^2-3x=-2\)

=>\(x^2-3x+2=0\)

=>(x-1)(x-2)=0

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

6: |x-5|-3=0

=>|x-5|=3

=>\(\left[\begin{array}{l}x-5=3\\ x-5=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=2\end{array}\right.\)

1) Ta có: \(4x+8=3x-1\)

\(\Leftrightarrow4x-3x=-1-8\)

\(\Leftrightarrow x=-9\)

2) Ta có: \(10-5\left(x+3\right)>3\left(x-1\right)\)

\(\Leftrightarrow10-5x-15-3x+3>0\)

\(\Leftrightarrow-8x>2\)

hay \(x< \dfrac{-1}{4}\)

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

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

a, \(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\left(ĐKXĐ:x\ne\pm2\right)\)

\(\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+2}+\frac{3}{x-2}\)

\(\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

Khử mẫu : \(9=\left(x-1\right)\left(x-2\right)+3\left(x+2\right)\)

Đến đây nhường bn, rất dễ =))

b, \(\frac{1}{x-5}-\frac{3}{x^2-6x+5}=\frac{5}{x-1}\)

\(\frac{1}{x-5}-\frac{3}{\left(x-5\right)\left(x-1\right)}=\frac{5}{\left(x-1\right)}\)

\(\frac{\left(x-1\right)}{x-5}-\frac{3}{\left(x-5\right)\left(x-1\right)}=\frac{5\left(x-5\right)}{\left(x-1\right)\left(x-5\right)}\)

Khử mẫu \(x-1-3=5\left(x-5\right)\)

Tự lm nốt mà cho mk hỏi, đề bài có bpt mà bpt đâu 

\(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\left(ĐKXĐ:x\ne2;-2\right)\)

\(< =>\frac{9}{x^2-2^2}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(< =>\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3x+6}{\left(x+2\right)\left(x-2\right)}\)

\(< =>9=x^2-2x-x+2+3x+6\)

\(< =>x^2-\left(2x+x-3x\right)+\left(2+6-9\right)=0\)

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

\(< =>x=\pm\sqrt{2}\left(tmđk\right)\)

Vậy tập nghiệm của phương trình trên là \(\pm\sqrt{2}\)

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):}`

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:

Đặt \(\hept{\begin{cases}S=x+y\\P=xy\end{cases}}\) hpt thành:

\(\hept{\begin{cases}S^2-P=3\\S+P=9\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S^2-P=3\\S=9-P\end{cases}}\Leftrightarrow\left(9-P\right)^2-P=3\)

\(\Leftrightarrow\orbr{\begin{cases}P=6\Rightarrow S=3\\P=13\Rightarrow S=-4\end{cases}}\).Thay 2 trường hợp S và P vào ta tìm dc

\(\hept{\begin{cases}x=3\\y=0\end{cases}}\)và\(\hept{\begin{cases}x=0\\y=3\end{cases}}\)

Câu 3: ĐK: \(x\ge0\)

Ta thấy \(x-\sqrt{x-1}=0\Rightarrow x=\sqrt{x-1}\Rightarrow x^2-x+1=0\) (Vô lý), vì thế \(x-\sqrt{x-1}\ne0.\)

Khi đó \(pt\Leftrightarrow\frac{3\left[x^2-\left(x-1\right)\right]}{x+\sqrt{x-1}}=x+\sqrt{x-1}\Rightarrow3\left(x-\sqrt{x-1}\right)=x+\sqrt{x-1}\)

\(\Rightarrow2x-4\sqrt{x-1}=0\)

Đặt \(\sqrt{x-1}=t\Rightarrow x=t^2+1\Rightarrow2\left(t^2+1\right)-4t=0\Rightarrow t=1\Rightarrow x=2\left(tm\right)\)

a, (4 - x )⁵ +(x - 2)⁵ =32

(=) 1024  -  x⁵ + x⁵  - 32 = 32

(=) -x⁵ + x⁵ = 32 + 32 - 1024

(=) 0x =  -960

=) phương trình vô nghiệm

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