Gọi số đường ở bao thứ nhất là x

\(\Rightarrow\) Số đường ở bao thứ 2 là : x - 16 ( kg)

Nếu chuyển từ bao thứ nhất sang bao thứ hai 10 kg , khi đó bao thứ nhất còn:

\(x-10\left(kg\right)\)

và bao thứ hai có:

\(x-16+10=x-6\) (kg)

Khi đó bao thứ hai sẽ nhiều hơn bao thứ nhất và nhiều hơn:

\(x-6-x-10=16\left(kg\right)\)

Sửa đề:

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

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

\(\Leftrightarrow x^2\left(x-1\right)+3x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+3x+6\right)=0\)

Nhận thấy : \(x^2+3x+6=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{15}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{15}{4}\ge0\forall x\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

vậy pt có tập nghiệm \(S=\left\{1\right\}\)

\(a,\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-48x^2+12x-1-64x^3-4x+48x^2+9\)\(=12x+8\)

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

\(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left[\left(x+y\right)^2-2xy\right]\)\(=2x^2-2xy+2y^2+6xy\) ( vì x + y =1 )

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

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

\(b,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\) \(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)\(\Leftrightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)

\(c,\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6x^2+12x+6-49=0\)\(\Leftrightarrow24x=-13\Rightarrow x=-\dfrac{13}{24}\)

\(d,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow-2x=23\Rightarrow x=-\dfrac{23}{2}\)

Sửa đề:

\(e,\left(x^2+1\right)^2-6\left(x^2+1\right)+9\)

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

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

\(P=27y^3+9y^2+y+\dfrac{1}{27}=\left(3y+3\right)^3\)

Với \(y=\dfrac{2}{3}\) ta có:

\(P=\left(3.\dfrac{2}{3}+3\right)^3=5^3=125\)

\(Q=x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x^2-2x+4xy\right)+4y^2-4y+10\)

\(=\left[x^2-2x\left(1-2y\right)+\left(1-2y\right)^2\right]+4y^2-4y+10-\left(1-2y\right)^2\)\(=\left(x+2y-1\right)^2+4y^2-4y+10-1+4y-4y^2\)\(=\left(x+2y-1\right)^2+9\)

Với \(x+2y=5\) , ta có:

\(Q=\left(5-1\right)^2+9=25\)

Đăng 1 lần thôi bạn-_-

\(x^4-104+13x^3-8x=0\)

\(\Leftrightarrow x^4+13x^3-8x-104=0\)

\(\Leftrightarrow x^3\left(x+13\right)-8\left(x+13\right)=0\)

\(\Leftrightarrow\left(x+13\right)\left(x^3-8\right)=0\)

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)\left(x^2+2x+4\right)=0\)

Ta có : \(x^2+2x+4=\left(x^2+2x+1\right)+3\)

\(=\left(x+1\right)^2+3\ge3\forall x\)

\(\Rightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)

\(K=\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right):\dfrac{2-4x}{x+1}-\dfrac{3x+1-x^2}{3x}\)\(=\left(\dfrac{\left(x+1\right)\left(x+2\right)}{3x\left(x+1\right)}+\dfrac{2.3x}{3x\left(x+1\right)}-\dfrac{3.3x\left(x+1\right)}{3x\left(x+1\right)}\right).\dfrac{x+1}{2\left(1-2x\right)}-1+\dfrac{1}{3x}-\dfrac{x}{3}\)\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}.\dfrac{x+1}{2\left(1-2x\right)}+\dfrac{1}{3x}-\dfrac{x}{3}-1\)\(=\dfrac{-2\left(4x^2-1\right)}{3x}.\dfrac{1}{-2\left(2x-1\right)}+\dfrac{1+x^2}{3x}-1\)

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

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

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

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

\(=-3x\left(x-1\right)\)

\(b,\left(x-3\right)^2-\left(x-3\right)\left(x^2+3x+9\right)\)

\(=x^3-9x^2+27x-27-x^3+27\)

\(=-9x\left(x+3\right)\)