a) / x + 5 / +3/ x - 2/ = / x + 4/ ( 1)

Lập bảng xét dấu , ta có :

*) Với : x < - 5 , ta có:

( 1 ) ⇔ - x - 5 + 3( 2 - x) = - x - 4

⇔ - x - 5 + 6 - 3x = - x - 4

⇔ 1 - 4x = -x - 4

⇔ 3x = 5

⇔ x = \(\dfrac{5}{3}\) ( không thỏa mãn )

*) Với : - 5 ≤ x < - 4 , ta có :

( 1) ⇔ x + 5 + 3( 2 - x ) = - x - 4

⇔ x + 5 + 6 - 3x = -x - 4

⇔ 11 - 2x = - x - 4

⇔ x = 15 ( không thỏa mãn )

*) Với : - 4 ≤ x < 2 , ta có :

( 1) ⇔ x + 5 + 3( 2 - x) = x + 4

⇔ x + 5 + 6 - 3x = x + 4

⇔ 11 - 2x = x + 4

⇔ 3x = 7

⇔ x = \(\dfrac{7}{3}\) ( không thỏa mãn )

*) Với : x ≥ 2 , ta có :

( 1) ⇔ x + 5 + 3( x - 2) = x + 4

⇔ x + 5 + 3x - 6 = x + 4

⇔ 4x - 1 = x + 4

⇔3x = 5

⇔ x = \(\dfrac{5}{3}\) ( không thỏa mãn )

Vậy , PT trên vô nghiệm

ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

=>(x-1)(x^3+4x^2+2x+3)=0

=>x-1=0

=>x=1

b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

ĐKXĐ: \(x\notin\left\{-3;1\right\}\)

Ta có: \(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)

\(\Leftrightarrow\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\frac{4}{\left(x-1\right)\left(x+3\right)}\)

Suy ra: \(\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)=4\)

\(\Leftrightarrow2x^2-2x-5x+5-2x^2-6x=4\)

\(\Leftrightarrow-13x+5=4\)

\(\Leftrightarrow-13x=4-5=-1\)

hay \(x=\frac{1}{13}\)(nhận)

Vậy: \(S=\left\{\frac{1}{13}\right\}\)

b) Có \(\left|2x+1\right|\ge0;\left|4x^2-1\right|\ge0\forall x\)

\(\Rightarrow\left|2x+1\right|+\left|4x^2-1\right|\ge0\forall x\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}2x+1=0\\4x^2-1=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(\left|2x-1\right|=\left|x+5\right|\) 

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+5\\2x-1=-x-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-\dfrac{4}{3}\end{matrix}\right.\)

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