(4x - 3)^3 + (3x - 2)^3 = (7x - 5)^3
<=> (4x - 3)^3 + (3x - 2)^3 - (7x - 5)^3 = 0
<=> (4x - 3 + 3x - 2 - 7x + 5)^3 = 0
<=> 4x - 3 + 3x - 2 - 7x + 5 = 0
<=> 4x + 3x - 7x = -5 + 2 + 3
<=> 0x = 0
Vậy pt có vô số nghiệm

nhớ k nhé

CM: Với a+b+c=0 thì \(a^3+b^3+c^3=3abc\)

Thật vậy: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow a^3=b^3+c^3=3abc\)

Áp dụng vào ta có:

\(PT\Leftrightarrow\left(4x-3\right)^3+\left(3x-2\right)^3+\left(5-7x\right)^3=0\)

Vì \(\left(4x-3\right)+\left(3x-2\right)+\left(5-7x\right)=0\)

\(\Rightarrow3\left(4x-3\right)\left(3x-2\right)\left(5-7x\right)=0\)

\(\Rightarrow x\in\left\{\frac{3}{4};\frac{2}{3};\frac{5}{7}\right\}\)

a) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)

\(\Leftrightarrow\left(\frac{x-45}{55}-1\right)+\left(\frac{x-47}{53}-1\right)=\left(\frac{x-55}{45}-1\right)+\left(\frac{x-53}{47}-1\right)\)

\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)

\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)

Vì \(\hept{\begin{cases}\frac{1}{55}< \frac{1}{45}\\\frac{1}{53}< \frac{1}{47}\end{cases}}\Rightarrow\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}< 0\)

\(\Rightarrow x-100=0\Rightarrow x=100\)

Vậy x = 100

Các phần sau tương tự nhé bạn

Ta có: \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)

\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)

\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)

Vì \(\hept{\begin{cases}\frac{1}{24}>\frac{1}{26}\\\frac{1}{25}>\frac{1}{27}\end{cases}}\Rightarrow\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\)

\(\Rightarrow x-23=0\Leftrightarrow x=23\)

Vậy x = 23

Ta có \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)

<=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)

<=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)

<=> x - 23 = 0 (Vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\ne0\))

<=> x = 23

Vậy x = 23 là nghiệm phương trình

Ta có:\(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)

\(\Leftrightarrow\left[\left(x^2+x\right)^2-2\left(x^2+x\right)\right]-\left[6\left(x^2+x\right)-12\right]=0\)

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

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

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

\(\Rightarrow x\in\left\{-3;-2;1;2\right\}\)

\(\left(x^2+x\right)^2+4.\left(x^2+x\right)-12=0\)

Đặt \(x^2+x=a\)ta có :

\(a^2+4a-12=0\)

\(\Leftrightarrow a^2+6a-2a-12=0\)

\(\Leftrightarrow a.\left(a+6\right)-2.\left(a+6\right)=0\)

\(\Leftrightarrow\left(a+6\right).\left(a-2\right)=0\)

Thay \(a=x^2+x\)vào phương trình trên ta có :

\(\left(x^2+x+6\right).\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+x+6=0\\x^2+x-2=0\end{cases}}\)

\(\Rightarrow x^2+x-2=0\)( vì \(x^2+x+6=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{23}{4}=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0\))

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

\(\Leftrightarrow x.\left(x-1\right)+2.\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right).\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{1;-2\right\}\)

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 \(P=\frac{-3}{2}\)