Ta có: \(\left(3x-1\right)^2-3\left(3x-2\right)=9\left(x+1\right)\left(x-3\right)\)

\(\Leftrightarrow9x^2-6x+1-9x+6=9\left(x^2-2x-3\right)\)

\(\Leftrightarrow9x^2-15x+7=9x^2-18x-27\)

\(\Leftrightarrow3x=-34\)

\(\Rightarrow x=-\frac{34}{3}\)

Ta có: ${\left(3x-1\right)}^2-3\left(3x-2\right)=9\left(x+1\right)\left(x-3\right)$

$\iff 9x^2-6x+1-9x+6=9\left(x^2-3x+x-3\right)$

$\iff 9x^2-15x+7=9x^2-18x-27$

$\iff 9x^2-15x+7-9x^2+18x+27=0$

$\iff 3x+34=0$

$\iff 3x=-34$

$\iff x=-\frac{34}{3}$

Vậy: $S=\left\{-\frac{34}{3}\right\}$

ĐK : a³ + a + 1 \(\ge0\)

Khi đó |a³ - a + 1| = a³ + a + 1

<=> \(\orbr{\begin{cases}a^3-a-1=a^3+a+1\\a^3-a-1=-a^3-a-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}2a+2=0\\2a^3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=-1\\a=0\end{cases}}\)

Khi a = 0 => a³ + a + 1 = 1 \(\ge\)0 (tm)

Khi a = -1 => a³ + a + 1 = -1 < 0 (loại)

Vậy a = 0 là nghiệm phương trình

5^9009 chia hết cho 5

suy ra 5^9009 có ít nhất 3 ước

mà số nguyên tố chỉ có nhiều nhất 2 ước

vậy 5^9009 ko là số nguyên tố

Hiển nhiên mà bạn

Ta có thể kể hàng loạt các ước số của \(5^{9009}\) như:

\(5;5^2;5^3;....;5^{9008};5^{9009}\)

=> \(5^{9009}\) không phải là số nguyên tố

Ta có : x² - 2x + 3 = ( x² - 2x + 1 ) + 2 = ( x - 1 )² + 2 ≥ 2 ∀ x

=> \(\frac{1}{x^2-2x+3}\le\frac{1}{2}\)

=> \(\frac{3}{x^2-2x+3}\le\frac{3}{2}\)

hay M ≤ 3/2

Đẳng thức xảy ra khi x = 1

Vậy MaxM = 3/2

x + y + z = 0 => x + y = -z <=> (x + y)² = (-z^2) <=> x² + y² + 2xy = z² <=> x² + y² - z² = -2xy

CMTT: y² + z² - x² = -2yz

z² + x² - y² = -2xz

Khi đó, ta có: A = \(\frac{1}{-2yz}+\frac{1}{-2xy}+\frac{1}{-2xz}\)

A = \(-\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}\right)=-\frac{1}{2}\cdot\frac{x+y+z}{xyz}=-\frac{1}{2}.0=0\)

Ta có: \(x+y+z=0\Rightarrow y+z=-x\Leftrightarrow\left(y+z\right)^2=\left(-x\right)^2\)

\(\Rightarrow y^2+z^2-x^2=-2yz\Rightarrow\frac{x^2}{y^2+z^2-x^2}=\frac{x^2}{-2yz}\)

Tương tự ta có: \(\frac{y^2}{z^2+x^2-y^2}=\frac{y^2}{-2xz};\frac{z^2}{x^2+y^2-z^2}=\frac{z^2}{-2xy}\)

\(\Rightarrow P=\frac{x^2}{-2yz}+\frac{y^2}{-2xz}+\frac{z^2}{-2xy}=\frac{x^3+y^3+z^3}{-2xyz}\)

\(=\frac{\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)}{-2xyz}\)

\(=\frac{0-3\left(-z\right)\left(-x\right)\left(-y\right)}{-2xyz}=\frac{3xyz}{-2xyz}=\frac{-3}{2}\)

Vậy \(P=\frac{-3}{2}\)

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

\(\Leftrightarrow2\left(x-3\right)\left(x+3\right)+\left(3x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-6+3x-4\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(5x-10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\5x-10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

b) Vì \(\left(4x+3\right)+\left(5-7x\right)+\left(3x-8\right)=0\)

\(\Rightarrow3\left(4x+3\right)\left(5-7x\right)\left(3x-8\right)=0\)

\(\Rightarrow x\in\left\{-\frac{3}{4};\frac{5}{7};\frac{8}{3}\right\}\)

mình ko hiểu câu b cho lắm

bạn giải thích kĩ hơn đc ko

ĐKXĐ : \(\hept{\begin{cases}ab-2\ne0\\ab+2\ne0\\a^4b^4\ne0\end{cases}}\Rightarrow ab\ne\pm2;a\ne0;b\ne0\)

\(P=\left(\frac{1}{ab-2}+\frac{1}{ab+2}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)

\(=\left(\frac{2ab}{a^2b^2-4}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)

\(=\left(\frac{4a^3b^3}{a^4b^4-16}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)

\(=\frac{8a^5b^5}{a^8b^8-16^2}.\frac{a^4b^4+16}{a^4b^4}=\frac{8a^5b^5\left(a^4b^4+16\right)}{\left(a^4b^4-16\right)\left(a^4b^4+16\right).a^4b^4}\)

\(=\frac{8ab}{a^4b^4-16}\)

b) Khi \(\frac{a^2+4}{b^2+9}=\frac{a^2}{9}\)

=> (a² + 4).9 = a²(b² + 9)

=> 9a² + 36 = a²b² + 9a²

=> a²b² = 36

=> (ab)² = 36

=> \(\orbr{\begin{cases}ab=6\left(tm\right)\\ab=-6\left(tm\right)\end{cases}}\)

Khi ab = 6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.6}{6^4-16}=\frac{48}{1280}=\frac{3}{80}\)

Khi ab = -6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.\left(-6\right)}{\left(-6\right)^4-16}=-\frac{3}{80}\)

Ta có: \(P=\frac{x+y}{xyz}=\frac{1}{yz}+\frac{1}{zx}\ge\frac{4}{yz+zx}\) (BĐT Cauchy-Schwarz)

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

\(=\frac{4}{-\left(z-\frac{1}{2}\right)^2+\frac{1}{4}}\ge\frac{4}{\frac{1}{4}}=16\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=y\\\left(z-\frac{1}{2}\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{4}\\z=\frac{1}{2}\end{cases}}\)

Vậy Min(P) = 16 khi \(\hept{\begin{cases}x=y=\frac{1}{4}\\z=\frac{1}{2}\end{cases}}\)

mình chưa học BĐT Cauchy nên ko hiểu bài cho lắm 

Vì \(x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1>0\left(\forall x\right)\)

\(\Rightarrow8x-4=0\Leftrightarrow x=\frac{1}{2}\)

Vậy x = 1/2

(8x - 4)(x² + 2x + 2) = 0

<=> \(\orbr{\begin{cases}8x-4=0\\x^2+2x+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x\in\varnothing\end{cases}}\Leftrightarrow x=\frac{1}{2}\)

Vậy x = 1/2 là nghiệm phương trình