a) Ta có: \(\left(\frac{1}{3}+2x\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)-\left(8x^3-\frac{1}{27}\right)\)

\(=\left(2x\right)^3+\left(\frac{1}{3}\right)^3-8x^3+\frac{1}{27}\)

\(=8x^3+\frac{1}{27}-8x^3+\frac{1}{27}\)

\(=\frac{2}{27}\)

Vậy: Giá trị của biểu thức \(\left(\frac{1}{3}+2x\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)-\left(8x^3-\frac{1}{27}\right)\) không phụ thuộc vào biến

b) Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x\)

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

\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)

\(=0\)

Vậy: Giá trị của biểu thức \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x\) không phụ thuộc vào biến

c) Ta có: \(y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)

\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)

\(=yx^4-y^5-yx^4+y^5\)

\(=0\)

Vậy: Giá trị của biểu thức \(y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\) không phụ thuộc vào biến

\(A=\sqrt{\left(x+1\right)^4+1}+\sqrt{\left(y-2\right)^4+1}\)

Đặt \(\left(x+1;y-2\right)=\left(a;b\right)\)

\(\Rightarrow\left(a+1\right)\left(b+1\right)=\frac{9}{4}\)

\(\Leftrightarrow ab+a+b=\frac{5}{4}\)

\(\Rightarrow\frac{a^2+b^2}{2}+\sqrt{2\left(a^2+b^2\right)}\ge\frac{5}{4}\)

\(\Rightarrow a^2+b^2\ge\frac{1}{2}\)

\(A=\sqrt{a^4+1}+\sqrt{b^4+1}\ge\sqrt{\left(a^2+b^2\right)^2+4}\ge\sqrt{\frac{1}{4}+4}=\frac{\sqrt{17}}{2}\)

Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\) hay \(\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=\frac{5}{2}\end{matrix}\right.\)

Câu 1 :

\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)=\left(2x\right)^3+y^3=8x^3+y^3\)Câu 2:

\(A=3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)\(\Leftrightarrow3\left(6x^2-2x-6\right)-2\left(4x^2+13x-12\right)+36x-9x^2=0\)\(\Leftrightarrow18x^2-6x-18-8x^2-26x+24+36x-9x^2=0\)\(\Leftrightarrow x^2+4x+6=0\)

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

Ta có:

\(\left(x+2\right)^2\ge0\forall x\)

Vậy pt vô nghiệm

Vậy:ko......

Câu 3:

\(\left(5x-3\right)\left(7x+2\right)-35x\left(x-1\right)=42\)

\(\Leftrightarrow35x^2+10x-21x-6-35x^2+35x-42=0\)\(\Leftrightarrow14x=48\Leftrightarrow x=\dfrac{7}{24}\)

Câu 4:

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

\(\Leftrightarrow6x^2-3x+10x-5+5x+10-6x^2-12x-x=0\)\(\Leftrightarrow-x=-5\Rightarrow x=5\)

câu 6,

Câu 6: \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(\Rightarrow10x^2+9x-\left(10x^2-2x+15x-3\right)=8\)

\(\Rightarrow10x^2+9x-10x^2+2x-15x+3=8\)

\(\Rightarrow-4x+3=8\)

\(\Rightarrow-4x=5\Rightarrow x=\dfrac{-5}{4}\)

Câu 7: \(x\left(x+1\right)\left(x+6\right)-x^3=5x\)

\(\Rightarrow\left(x^2+x\right)\left(x+6\right)-x^3=5x\)

\(\Rightarrow x^3+x^2+6x^2+6x-x^3=5x\)

\(\Rightarrow7x^2=-x\)

\(\Rightarrow7x=-1\Rightarrow x=\dfrac{-1}{7}\).

Câu 2: \(A=3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)\)

\(=\left(6x-9\right)\left(3x+2\right)-\left(2x+8\right)\left(4x-3\right)+36x-9x^2\)

\(=18x^2-27x+12x-18-8x^2-32x+6x+24+36x-9x^2\)

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

\(=x^2-2x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}+ 6\)

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

Khi đó: \(\left(x-\dfrac{5}{2}\right)^2-\dfrac{1}{4}=0\)

\(\Rightarrow\left(x-\dfrac{5}{2}\right)^2=\left(\dfrac{1}{2}\right)^2\)

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

a/ Ta có 

\(K^4+\frac{1}{4}=K^4+K^2+\frac{1}{4}-K^2=\left(K^2+\frac{1}{2}\right)^2-K^2=\left(K^2+K+\frac{1}{2}\right)\left(K^2-K+\frac{1}{2}\right)\)

Ta lại có 

\(K^2+K+\frac{1}{2}=\left(K+1\right)^2-\left(K+1\right)+\frac{1}{2}\)

\(\Rightarrow K^4+\frac{1}{4}=\left(K^2-K+\frac{1}{2}\right)\left(\left(K+1\right)^2-\left(K+1\right)+\frac{1}{2}\right)\)

Áp dụng vào bài toán ta được

\(=\frac{101^2-101+0,5}{1^2-1+0,5}=20201\)\(1S=\frac{\left(2^2-2+0,5\right)\left(3^2-3+0,5\right)\left(4^2-4+0,5\right)\left(5^2-5+0,5\right)...\left(100^2-100+0,5\right)\left(101^2-101+0,5\right)}{\left(1^2-1+0,5\right)\left(2^2-2+0,5\right)\left(3^2-3+0,5\right)\left(4^2-4+0,5\right)...\left(99^2-99+0,5\right)\left(100^2-100+0,5\right)}\)

b/

\(\frac{3\left(x+y\right)}{3\sqrt{x\left(4x+5y\right)}+3\sqrt{y\left(4y+5x\right)}}\)

\(\ge\frac{3\left(x+y\right)}{\frac{9x+4x+5y}{2}+\frac{9y+4y+5x}{2}}\)

\(=\frac{1}{3}\)

Dấu = xảy ra khi x = y

Bấm sao mà nói đẩy đáp số lên trên mất rồi

\(\Rightarrow1S=\frac{101^2-101+0,5}{1^2-1+0,5}=20201\)

câu hệ sao từ x^7-y^14 sao xuống đc (x-y^2)(x+y^2) ?