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

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

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

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

\(M=279\)

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:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

=>3x-9-10x+2=-4

=>-7x-7=-4

=>-7x=3

=>x=-3/7

b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)

=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)

=>10-2x+7x-14=4x-4+x

=>5x-4=5x-4

=>0x=0(luôn đúng)

Vậy: S=R\{0;2}

giup minh voi cac bạn

\(\Rightarrow2x\cdot\left(64x^2-16x+1\right)\cdot\left(4x-1\right)=9\)

\(\Rightarrow\left(64x^2-16x+1\right)\cdot\left(8x^2-2x\right)=9\)

Nhân cả hai vế của phương trình với 8 ta được:

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

Đặt \(a=64x^2-16x\left(a\ge1\right)\) (1)

\(\Rightarrow\left(a+1\right)\cdot a=72\)

\(\Rightarrow a^2+a-72=0\)

\(\Rightarrow\left(a-8\right)\cdot\left(a+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=8\left(tmđk\right)\\a=-9\left(loại\right)\end{matrix}\right.\)

Thay vào (1) ta đc:

\(64x^2-16x=8\Rightarrow64x^2-16x-8=0\)

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

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

$2x{}^{}\left(4x-1\right)=9$

$\iff \left(64{}^{}-16x+1\right)\left(64{}^{}-16x\right)=72$

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

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

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

⇔[ t=−9
t=8
⇔[t=−9(loai)t=8(nhan

$\Rightarrow 64{}^{}-16$

Làm cho bạn 1 con thôi dài quá trôi hết màn hình:

c) có vẻ khó nhất (con khác tương tự)

đặt 2x+2=t=> x+1=t/2

\(\left(t-1\right).\left(\frac{t}{2}\right)^{^2}.\left(t+1\right)=18\Leftrightarrow\left(t^2-1\right)t^2=4.18\)

\(t^4-t^2=4.18\Leftrightarrow y^2-2.\frac{1}{2}y+\frac{1}{4}=4.18+\frac{1}{4}=\frac{16.18+1}{4}=\left(\frac{17}{2}\right)^2\)

<=> \(\left(y-\frac{1}{2}\right)^{^2}=\left(\frac{17}{2}\right)^2\Rightarrow\left[\begin{matrix}y=\frac{1}{2}-\frac{17}{2}=-8\\y=\frac{1}{2}+\frac{17}{2}=9\end{matrix}\right.\Rightarrow\left[\begin{matrix}2x+2=-8\Rightarrow x=-5\\2x+2=9\Rightarrow x=\frac{7}{2}\end{matrix}\right.\)

\(\frac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)

\(=\frac{x^2\left(3x^2-2x+1\right)-2x\left(3x^2-2x+1\right)-5\left(3x^2-2x+1\right)}{3x^2-2x+1}\)

\(=\frac{\left(3x^2-2x+1\right)\cdot\left(x^2-2x-5\right)}{3x^2-2x+1}\)

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

\(\frac{2x^3-9x^2+19x-15}{x^2-3x+5}\)

\(=\frac{2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)}{x^2-3x+5}\)

\(=\frac{\left(x^2-3x+5\right)\left(2x-3\right)}{x^2-3x+5}\)

\(=2x-3\)

\(\frac{\left(8x^3-y^3\right)\left(4x^2-y^2\right)}{\left(2x+y\right)\left(4x^2-4xy+y^2\right)}\)

\(=\frac{\left(2x-y\right)\left(4x^2+2xy+y^2\right)\left(2x-y\right)\left(2x+y\right)}{\left(2x+y\right)\left(2x-y\right)^2}\)

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

Ta có: \(2x\left(8x-1\right)^2\cdot\left(4x-1\right)=9\)

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

\(\Leftrightarrow\left(64x^2-16x+1\right)\left(8x^2-2x\right)-9=0\)

\(\Leftrightarrow512x^4-128x^3-128x^3+32x^2+8x^2-2x-9=0\)

\(\Leftrightarrow512x^4-256x^3+40x^2-2x-9=0\)

\(\Leftrightarrow256x^3\left(2x-1\right)+40x^2-20x+18x-9=0\)

\(\Leftrightarrow256x^3\left(2x-1\right)+20x\left(2x-1\right)+9\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(256x^3+20x+9\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(256x^3+64x^2-64x^2-16x+36x+9\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left[64x^2\left(4x+1\right)-4x\left(4x+1\right)+9\left(4x+1\right)\right]=0\)

\(\Leftrightarrow\left(2x-1\right)\left(4x+1\right)\left(64x^2-4x+9\right)=0\)

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

Vậy: \(x\in\left\{\dfrac{1}{2};-\dfrac{1}{4}\right\}\)