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Bài 2 :

a, Ta có : \(x^2-5x+4< 0\)

\(\Leftrightarrow x^2-x-4x+4< 0\)

\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)< 0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1\right)< 0\)

Vậy ...

b, Ta có : \(\dfrac{x-3}{x+1}< 1\)

\(\Leftrightarrow\dfrac{x-3}{x+1}-\dfrac{x+1}{x+1}< 0\)

\(\Leftrightarrow\dfrac{x-3-x-1}{x+1}=\dfrac{-4}{x+1}< 0\)

Thấy - 4 < 0

Nên để \(-\dfrac{4}{x+1}< 0\) <=> x + 1 > 0 ( TH A, B trái dấu )

Vậy ...

Bài 1: 

a) ĐKXĐ: \(x\ge2\)

Ta có: \(3\sqrt{x-2}+\sqrt{25x-50}=2^5\)

\(\Leftrightarrow3\sqrt{x-2}+5\sqrt{x-2}=32\)

\(\Leftrightarrow8\sqrt{x-2}=32\)

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

\(\Leftrightarrow x-2=16\)

hay x=18(thỏa ĐK)

Vậy: S={18}

\(pt\Leftrightarrow4x-20-3x+1=x\Leftrightarrow4x-x-3x=-1+20\Leftrightarrow0x=19\left(vl\right)\)

⇒ pt vô nghiệm

Ta có 4(x-5)-(3x-1)=x

\(\Leftrightarrow4x-20-3x+1-x=0\)

\(\Leftrightarrow-19\ne0\)

Vậy: \(x\in\varnothing\)

\(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}\)

⇔ \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=\dfrac{4x+2}{7}\)

⇔ \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=\dfrac{4x+2}{7}\)

⇔ \(\dfrac{140x-84}{168}-\dfrac{294x-42}{168}=\dfrac{96x+48}{168}\)

⇔ 140x-84-294x+42=96x+48

⇔ -154x-42=96x+48

⇔ -250x=90

⇔ x=\(\dfrac{-9}{26}\)

Vậy phương trình đã cho có tập nghiệm S={\(\dfrac{-9}{26}\)}

bạn đăng tách cho mn cùng giúp nhé 

Bài 1 : 

a, \(\Leftrightarrow11-x=12-8x\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

b, \(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

\(\Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\Leftrightarrow x=-2\)

c, \(\Leftrightarrow3-2x=-x-4\Leftrightarrow x=7\)

d, \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)

\(\Leftrightarrow3x^2+12x-9=3x^2+3x+1\Leftrightarrow x=\dfrac{10}{9}\)

e, \(\Leftrightarrow2x^2-x-3=2x^2+9x-5\Leftrightarrow x=5\)

f, \(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-22\)

\(\Leftrightarrow-5x^2+2x-1=-5x^2-x-22\Leftrightarrow3x=-21\Leftrightarrow x=-7\)

h) \(PT\Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\)

\(\Leftrightarrow3x=24\)

\(\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

j) \(PT\Leftrightarrow x^3-x^2+x+x^2-x+1-2x=x^3-x\)

\(\Leftrightarrow x=1\)

Vậy: \(S=\left\{1\right\}\)

Tham khảo: 

\(x=\dfrac{1}{a}.\sqrt{\dfrac{2a}{b}-1}\Rightarrow ax=\sqrt{\dfrac{2a}{b}-1}\)

\(\Rightarrow\left\{{}\begin{matrix}1+ax=\dfrac{\sqrt{2a-b}+\sqrt{b}}{\sqrt{b}}\\1-ax=\dfrac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1-ax}{1+ax}=\dfrac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}+\sqrt{2a-b}}=\dfrac{\left(\sqrt{b}-\sqrt{2a-b}\right)^2}{2\left(b-a\right)}\)

Lại có:

\(\dfrac{1+bx}{1-bx}=\dfrac{a+\sqrt{2ab-b^2}}{a-\sqrt{2ab-b^2}}=\dfrac{a^2-\left(2ab-b^2\right)}{\left(a-\sqrt{2ab-b^2}\right)^2}=\dfrac{\left(a-b\right)^2}{\left(a-\sqrt{2ab-b^2}\right)^2}\)

\(\Rightarrow\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{b-a}{a-\sqrt{2ab-b^2}}\)

\(\Rightarrow A=\dfrac{1-ax}{1+ax}.\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{\left(\sqrt{b}-\sqrt{2a-b}\right)^2}{2a-2\sqrt{2ab-b^2}}=\dfrac{2a-2\sqrt{2ab-b^2}}{2a-2\sqrt{2ab-b^2}}=1\)

a: \(\Leftrightarrow\left(x-3\right)\left[m\left(x+3\right)-1\right]=0\)

=>x=3 hoặc m(x+3)=1

b: \(\Leftrightarrow m^2x-m-x\left(3m-2\right)+1=0\)

\(\Leftrightarrow x\left(m^2-3m+2\right)=m-1\)

\(\Leftrightarrow x\left(m-1\right)\left(m-2\right)=m-1\)

Để phương trình vô nghiệm thì m-2=0

hay m=2

Để phương trình có vô số nghiệm thì m-1=0

hay m=1

Để phương trình có nghiệm duy nhất thì (m-1)(m-2)<>0

hay \(m\notin\left\{1;2\right\}\)