Giups mik vs

nhân phá tung ra là xog mà

\(M=\left(7-2x\right)\left(4x^2+14x+49\right)-\left(64-8x^3\right)\)

\(M=\left(7-2x\right)\left[\left(2x\right)^2+2x\cdot7+7^2\right]-\left(64-8x^3\right)\)

\(M=\left[7^3-\left(2x\right)^3\right]-\left(64-8x^3\right)\)

\(M=343-8x^3-64+8x^3\)

\(M=279\)

Vậy M có giá trị 279 với mọi x

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

\(P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3\)

\(P=16x^3-8x^2+4x-2\)

Thay \(x=10\) vào P ta có:

\(P=16\cdot10^3-8\cdot10^2+4\cdot10-2=15238\)

Vậy P có giá trị 15238 tại x=10

a: M=343-8x^3-64+8x^3=279

b: P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3

=16x^3-8x^2+4x-2

=16*10^3-8*10^2+4*10-2=15238

a ) \(x\left(x+1\right)\left(x^2+x+1\right)=42\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)=42\)

Đặt \(x^2+x=t\), ta được :

\(t\left(t+1\right)=42\)

\(\Leftrightarrow t^2+t-42=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=6\\t=-7\end{matrix}\right.\)

Khi t = 6, ta được :

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Khi t = -7, ta được :

\(x^2+x+7=0\)

\(\Leftrightarrow\left[x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{27}{4}=0\) ( Vô lí )

Vậy ...

Câu a:

\(2x\left(8x-1\right)^2\left(4x-1\right)=9\)

\(\Leftrightarrow\left(64x^2-16x+1\right)\left(64x^2-16x\right)=72\)

Đặt 64x² - 16x = t \(\left(t\ge-1\right)\)

\(\Rightarrow t\left(t+1\right)=72\)

\(\Leftrightarrow\left(t+9\right)\left(t-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-9\left(loai\right)\\t=8\left(nhan\right)\end{matrix}\right.\)

\(\Rightarrow64x^2-16x=8\)

\(\Leftrightarrow8\left(2x-1\right)\left(4x+1\right)=0\)

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

Câu b:

\(\Leftrightarrow\left(x+1\right)^2\left(2x+1\right)\left(2x+3\right)=18\)

\(\Leftrightarrow\left(4x^2+8x+4\right)\left(4x^2+8x+3\right)=72\)

Đặt 4x² + 8x + 4 = m \(\left(m\ge0\right)\)

\(\Rightarrow m\left(m-1\right)=72\)

\(\Leftrightarrow\left(m-9\right)\left(m+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=9\left(nhan\right)\\m=-8\left(loai\right)\end{matrix}\right.\)

\(\Rightarrow4\left(x+1\right)^2=9\)

\(\Leftrightarrow x+1=\pm\dfrac{3}{2}\)

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

d: Ta có: \(4x\left(2x+3\right)-8x\left(x+4\right)\)

\(=8x^2+12x-8x^2-32x\)

=-20x

e: Ta có: \(2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)

\(=10x^2+4x+6x^2-2x-9x+3\)

\(=16x^2-7x+3\)

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

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

\(=4x^2+x-4\)

=>x=1