\(3x\left(2x-4\right)-6x\left(x+5\right)=x-1\)

\(\Leftrightarrow6x^2-12x-6x^2-30x-x+1=0\)

\(\Leftrightarrow-43x+1=0\)

\(\Leftrightarrow x=\dfrac{1}{43}\)

Bài tiếp:

\(a)-10x+5x\left(x^2-7x+2x^2\right)-x^2\left(5x-8\right)+27x^2\)

\(=-10x+5x^3-35x^2+10x^3-5x^3+8x^2+27x^2\)

\(=10x^3-10x\)

Vậy biểu thức phụ thuộc vào biến \(x\).

\(b)4x\left(3x+5\right)-6x\left(2x-3\right)-38x+5\)

\(=12x^2+20x-12x^2+18x-38x+5\)

\(=5\)

Vậy biểu thức không phụ thuộc vào biến \(x\).

<=> 6x² - 12x - 6x² - 30x = x-1

<=> -32x = x-1 

<=> -33x=-1

<=> x=\(\dfrac{1}{33}\)

a, x² + 2x + 1 = (x+1)(x+1)

b,   9x²  + 6xy + y² = (3x + y)(3x+y)

c,     x⁴ -  2x² + 1 =  (x² - 1)(x² -1)

d, 9x² + 16 - 24 x = (3x - 4)(3x -4)

a, x2 + 2x + 1 = (x+1)(x+1)

b,   9x2  + 6xy + y2 = (3x + y)(3x+y)

c,     x4 -  2x2 + 1 =  (x2 - 1)(x2 -1)

d, 9x2 + 16 - 24 x = (3x - 4)(3x -4)

\(\dfrac{3x^2-3x+3}{x^3+1}\)

\(=\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{3}{x+1}\)

a) \(=\left(x+1\right)^2\)

b) \(=\left(3x+y\right)^2\)

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

d) \(=\left(3x-4\right)^2\)

\(a,\left(x+3\right)^3=x^3+9x^2+27x+27\)

\(b,\left(x+\dfrac{1}{3}\right)^3=x^3+x^2+\dfrac{1}{3}x+\dfrac{1}{27}\)

\(c,\left(2x-y\right)^3=8x^3-12x^2y+6xy^2-y^3\)

\(d,\left(x-\dfrac{1}{2}\right)^3=x^3-\dfrac{3}{2}x^2+\dfrac{3}{4}x-\dfrac{1}{8}\)

câu hỏi là j?

Do you know anything about the Bunyakovsky's inequality? It states that:

"With 2 sets of numbers \(\left(a_1,a_2,a_3,...,a_n\right)\) and \(\left(b_1,b_2,b_3,...,b_n\right)\), we have \(\left(a_1^2+a_2^2+a_3^2+...+a_n^2\right)\left(b_1^2+b_2^2+b_3^2+...+b_n^2\right)\)\(\ge\left(a_1b_1+a_2b_2+a_3b_3+...+a_nb_n\right)^2\)."

If you want to study more about this inequality, please check it on the Internet. Now, I'll give you the summary solution:

We have \(\left(a^2+b^2+c^2+d^2\right)\left(1^2+1^2+1^2+1^2\right)\)\(\ge\left(a.1+b.1+c.1+d.1\right)^2\) 

\(\Leftrightarrow4\left(a^2+b^2+c^2+d^2\right)\ge4\) (Because \(a+b+c+d=2\))

\(\Leftrightarrow a^2+b^2+c^2+d^2\ge1\)

"=" happens when \(a=b=c=d=\dfrac{1}{2}\)

Áp dụng BĐT Caushy ta có:

\(A^2+\dfrac{1}{4}\ge A;B^2+\dfrac{1}{4}\ge B;C^2+\dfrac{1}{4}\ge C;D^2+\dfrac{1}{4}\ge D\)

\(\Rightarrow A^2+B^2+C^2+D^2+1\ge A+B+C+D=2\)

\(\Leftrightarrow A^2+B^2+C^2+D^2\ge1\left(đpcm\right)\)

Dấu "=" xảy ra \(\Leftrightarrow A=B=C=D=1\)

\(x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)

\(=x^4-xy+x^2y-x^4-x^2y+3xy\)

\(=2xy\)

Thay \(x=\dfrac{1}{4};y=-2005\) vào, ta được:

\(2.\dfrac{1}{4}.\left(-2005\right)=\dfrac{1}{2}.\left(-2005\right)=-\dfrac{2005}{2}\)