Ta có: \(D=\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)

\(=\left(x-y+x+y\right)^2\)

\(=4x^2\)

Bạn quá dảnh nên bạn lám mỗi câu đó thôi à :)

A= x²+x-2-x+4

  =x²+2

Vì x² >=0 => x²+2>0

 Vậy pj]ơng trình vô nghiệm.

A= (x-1).(x+2)-(x-4)=0

^{x2+2X-X-2-X+4=0}

x2-2=0

x=\(\sqrt{2}\)

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

Vì |1/4 - x| ≥ 0; |x - y + z| ≥ 0; |2/3 + y| ≥ 0

=> |1/4 - x| + |x - y + z| + |2/3 + y| ≥ 0

Dấu " = " xảy ra <=>. \(\hept{\begin{cases}\frac{1}{4}-x=0\\x-y+z=0\\\frac{2}{3}+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\\frac{1}{4}-y-\frac{2}{3}=0\\y=\frac{-2}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=\frac{-5}{12}\\z=\frac{-2}{3}\end{cases}}\) 

Vậy ....

a. \(8x\left(x-2007\right)-2x+4034=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy x=2017 hoặc x=1/4

b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)

\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy x=0 hoặc x=-4

c.\(4-x=2\left(x-4\right)^2\)

\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)

\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy x=4 hoặc x=7/2

d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)

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

Nxet: (x²+3)>0 với mọi x

=> x-2=0 <=>x=2

Vậy x=2

a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0

     4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0

     4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0

     4\(x^2\) - 8029\(x\) + 2017 = 0

     4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))²) - (\(\dfrac{8029}{4}\))²  + 2017 = 0

    4.(\(x\) + \(\dfrac{8029}{8}\))² = (\(\dfrac{8029}{4}\))² - 2017

       \(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\) 

a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\)

\(\Leftrightarrow x^3+9x+2=x^3+8\)

\(\Leftrightarrow x^3+9x=x^3+8-2\)

\(\Leftrightarrow x^3+9x=x^3+6\)

\(\Leftrightarrow x^3+9x=x^3+6x-x^3\)

\(\Leftrightarrow\frac{2}{3}\)

b) \(x^2-4=8\left(x-2\right)\)

\(\Leftrightarrow x^2-4=8x-16\)

\(\Leftrightarrow x^4-4=8x-16+16\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

giúp mình bài ni với :3x^2(x+1)-5x(x+1)^2+4(x+1)

Ta có : x² + x + 1

= ( x² + x + 1/4 ) + 3/4

= ( x + 1/2 )² + 3/4 ≥ 3/4 ∀ x

Dấu "=" xảy ra khi x = -1/2

=> GTNN của biểu thức = 3/4 <=> x = -1/2

Đặt \(A=x^2+x+1\)  , ta có :

\(A=x^2+x+1\)

\(=x^2+x+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow minA=\frac{3}{4}\) khi và chỉ khi  \(x=\frac{-1}{2}\)