a) \(\left(x^2+x+1\right).\left(6-2x\right)=0\)

\(\Leftrightarrow6-2x=0\)( vì x² + x + 1 > 0 với mọi x )

\(\Leftrightarrow x=3\)

Vậy phương trình có 1 nghiệm x = 3

b) \(\left(8x-4\right).\left(x^2+2x+2\right)=0\)

\(\Leftrightarrow8x-4=0\)( vì x² + 2x + 2 > 0 với mọi x )

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy phương trình có 1 nghiệm \(x=\frac{1}{2}\)

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

\(\Leftrightarrow\left(2x^2+x-6\right)^2+3\left(2x^2+x-6\right)=0\)

\(\Leftrightarrow\left(2x^2+x-6\right)\left(2x^2+x-6+3\right)=0\)

\(\Leftrightarrow\left(2x^2+x-6\right)\left(2x^2+x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-3\right)\left(x-1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x-3=0\end{cases}}\)hoặc \(\orbr{\begin{cases}x-1=0\\2x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{3}{2}\end{cases}}\)hoặc \(\orbr{\begin{cases}x=1\\x-\frac{3}{2}\end{cases}}\)

Vậy tập nghiệm của PT là \(S=\left\{-2;\frac{3}{2};1;-\frac{3}{2}\right\}\)

b) \(2y^4-9y^3+14y^2-9y+2=0\)

\(\Leftrightarrow\left(y-2\right)\left(y-1\right)^2\left(2y-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y-2=0\\\left(y-1\right)^2=0\end{cases}}\)hoặc \(2y-1=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=2\\y-1=0\end{cases}}\)hoặc \(2y=1\)

\(\Leftrightarrow\orbr{\begin{cases}y=2\\y=1\end{cases}}\)hoặc \(y=\frac{1}{2}\)

Vậy tập nghiệm của PT là \(S=\left\{2;1;\frac{1}{2}\right\}\)

a) Đặt 2x² + x - 6 = a

pt <=> a² + 3( a + 3 ) - 9 = 0

<=> a² + 3a + 9 - 9 = 0

<=> a( a + 3 ) = 0

<=> ( 2x² + x - 6 )( 2x² + x - 6 + 3 ) = 0

<=> ( 2x² + x - 6 )( 2x² + x - 3 ) = 0

<=> ( 2x² + 4x - 3x - 6 )( 2x² - 2x + 3x - 3 ) = 0

<=> [ 2x( x + 2 ) - 3( x + 2 ) ][ 2x( x - 1 ) + 3( x - 1 ) ] = 0

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

<=> x = -2 hoặc x = 1 hoặc x = ±3/2

Vậy S = { -2 ; 1 ; ±3/2 }

b) 2y⁴ - 9y³ + 14y² - 9y + 2 = 0

<=> 2y⁴ - 4y³ - 5y³ + 10y² + 4y² - 8y - y + 2 = 0

<=> 2y³( y - 2 ) - 5y²( y - 2 ) + 4y( y - 2 ) - ( y - 2 ) = 0

<=> ( y - 2 )( 2y³ - 5y² + 4y - 1 ) = 0

<=> ( y - 2 )( 2y³ - 2y² - 3y² + 3y + y - 1 ) = 0

<=> ( y - 2 )[ 2y²( y - 1 ) - 3y( y - 1 ) + ( y - 1 ) ] = 0

<=> ( y - 2 )( y - 1 )( 2y² - 3y + 1 ) = 0

<=> ( y - 2 )( y - 1 )( 2y² - 2y - y + 1 ) = 0

<=> ( y - 2 )( y - 1 )[ 2y( y - 1 ) - ( y - 1 ) ] = 0

<=> ( y - 2 )( y - 1 )²( 2y - 1 ) = 0

<=> y = 2 hoặc y = 1 hoặc y = 1/2

Vậy S = { 2 ; 1 ; 1/2 }

\(0\le a,b,c\le1\)\(\Rightarrow1-a,1-b,1-c\ge0\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge0\)

\(\Rightarrow1+ab+bc+ca-\left(a+b+c\right)-abc\ge0\)

\(\Rightarrow1+ab+bc+ca\ge a+b+c+abc\left(1\right)\)

Bài toán : \(a+b^2+c^3-ab-bc-ca\le1\)

trở thành : \(1+ab+bc+ca\ge a+b^2+c^3\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\Rightarrow\)chỉ cần c/m \(a+b+c+abc\ge a+b^2+c^3\left(3\right)\)thì bài toán được chứng minh

\(\left(3\right)\Leftrightarrow b\left(1-b\right)+c\left(1-c^2\right)+abc\ge0\left(luon-dung\right)\)

Vậy bài toán được c/m

Dấu "=" xảy ra\(\orbr{\begin{cases}\left(a,b,c\right)-la-hoan-vi-cua-\left(0,0,1\right)\\\left(a,b,c\right)-la-hoan-vi-cua-\left(0,1,1\right)\end{cases}}\)

\(\frac{2x+3}{3}-\frac{x}{7}=\frac{5-x}{3}\)

<=> \(\frac{2x+3}{3}-\frac{x}{7}-\frac{5-x}{3}=0\)

<=> \(\frac{3x-2}{3}-\frac{x}{7}=0\)

<=> \(\frac{7\left(3x-2\right)-3x}{21}=0\)

<=> \(\frac{21x-14-3x}{21}=0\)

<=> 21x - 14 - 3x = 0

<=> 18x = 14

<=> x = 7/9 

Vậy x = 7/9 là nghiệm phương trình

\(\frac{2x+3}{3}-\frac{x}{7}=\frac{5-x}{3}\)

\(\Leftrightarrow\frac{14x+21}{21}-\frac{3x}{21}=\frac{35-7x}{21}\)

\(\Leftrightarrow\frac{11x+21}{21}=\frac{35-7x}{21}\)

\(\Leftrightarrow11x+21=35-7x\)

\(\Leftrightarrow18x=14\)

\(\Leftrightarrow x=\frac{7}{9}\)

Vậy...........

Trả lời:

       3x - 2 = 2x + 3

<=> 3x - 2x = 3 + 2

<=> x = 5

Vậy S = { 5 }

        5 - 4x = -x + 11

<=> -4x + x = 11 - 5

<=> -3x = 6

<=> x = -2

Vậy S = { -2 }

\(\frac{x}{x+1}-\frac{2x-3}{1-x}=\frac{3x^2+5}{x^2-1}\)ĐK : \(x\ne\pm1\)

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

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

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

\(\Leftrightarrow-2x-2=0\Leftrightarrow x=-1\) vô lí 

Vậy phương trình vô nghiệm 

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

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

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

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

\(\Leftrightarrow-2x-8=0\)

\(\Leftrightarrow-2x=8\)

\(\Leftrightarrow x=-4\)