Đặt t = x + 3 
=> x + 2 = t - 1; x + 4 = t + 1. 
ta có pt: (t - 1)^4 + (t + 1)^4 = 82 
<=>[(t -1)²]² + [(t + 1)²]² = 82 
<=> (t² - 2t + 1)² + (t² + 2t + 1)² = 82 
<=> (t²+1)² - 4t(t²+1) + 4t² + (t²+1)² + 4t(t²+1) + 4t² = 82 
<=> (t² + 1)² + 4t² = 41 
<=> t^4 + 6t² + 1 = 41 
<=> (t²)² + 6t² - 40 = 0 
<=> t² = -10 (loại) hoặc t² = 4 
<=> t = 2 hoặc t = -2 
với t = -2 => x = -5 
với t = 2 => x = -1 
vậy pt có hai nghiệm là : x = -1 hoặc x = -5

Giải thík hộ mk tại sao (t^2-2t+1)^2 lại bằng (t^2+1)^2

Viết lại đề đi!!

3)

\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)

có vấn đề thì phải

= 5 mới đúng chứ-.-

p/s: câu c = -5 chứ-.-

3)

\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)

\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=-5+5\)

\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)

\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)

\(\Leftrightarrow50-x=0\) Vì \(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\ne0\)

\(\Leftrightarrow x=50\)

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

\(\Leftrightarrow5x-15-\left(5x^2-11x+2\right)=5x^2\)

\(\Leftrightarrow5x-15-5x^2+11x-2=5x^2\)

\(\Leftrightarrow-10x^2+16x-17=0\)

\(\cdot\Delta=16^2-4.\left(-10\right).\left(-17\right)=-304< 0\)

Vậy pt vô nghiệm

a: \(5x-20x^2=0\)

\(\Leftrightarrow5x\left(1-4x\right)=0\)

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

c: \(x\left(x-3\right)-5x+15=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)

\(a,-3x^2+5x=0\)

\(\Rightarrow x\left(5-3x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\5-3x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy................

\(b,x^2-5x-24=0\)

\(\Rightarrow x^2-8x+3x-24=0\)

\(\Rightarrow\left(x^2+3x\right)-\left(8x+24\right)=0\)

\(\Rightarrow x\left(x+3\right)-8\left(x+3\right)=0\)

\(\Rightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

Vậy........................

=>5x^2-6x-11=0

=>5x^2-11x+5x-11=0

=>(5x-11)(x+1)=0

=>x=11/5 hoặc x=-1

1.

Đặt \(x^2-5x=a\Rightarrow a^2=\left(x^2-5x\right)^2\)

Thay vào pt:

\(\Rightarrow a^2+10a+24=0\)

\(\Leftrightarrow a^2+6a+4a+24=0\)

\(\Leftrightarrow a\left(a+6\right)+4\left(a+6\right)=0\)

\(\Leftrightarrow\left(a+6\right)\left(a+4\right)=0\)

\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)

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

\(\Leftrightarrow\left[x\left(x-3\right)-2\left(x-3\right)\right]\left[x\left(x-4\right)-\left(x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x-4\right)\left(x-1\right)=0\)

\(\Rightarrow x-3=0,x-2=0,x-4=0,x-1=0\)

\(\Rightarrow x=3,x=2,x=4,x=1\)

T I C K mình sẽ giải típ cho cảm ơn

típ nha

Danh ơi ! xem lại đề của bài 2 nha 

ĐKXĐ: \(x^3-1\ge0\Rightarrow\left(x-1\right)\left(x^2+x+1\right)\ge0\) 

mà \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

\(\Rightarrow x-1\ge0\Rightarrow x\ge1\)

\(2x^2+5x-1=7\sqrt{x^3-1}\Leftrightarrow2x^2+2x+2+3x-3=7\sqrt{x-1}\sqrt{x^2+x+1}\)

\(\Leftrightarrow2\left(x^2+x+1\right)+3\left(x-1\right)=7\sqrt{x-1}\sqrt{x^2+x+1}\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\\b=\sqrt{x^2+x+1}\end{matrix}\right.\left(a,b\ge0\right)\)

\(\Rightarrow\) pt trở thành \(2b^2+3a^2=7ab\Rightarrow2b^2-7ab+3a^2=0\)

\(\Rightarrow2b^2-6ab-ab+3a^2=0\Rightarrow2b\left(b-3a\right)-a\left(b-3a\right)=0\)

\(\Rightarrow\left(b-3a\right)\left(2b-a\right)=0\Rightarrow\left[{}\begin{matrix}b=3a\\2b=a\end{matrix}\right.\)

\(TH_1:b=3a\Rightarrow\sqrt{x^2+x+1}=3\sqrt{x-1}\)

\(\Rightarrow x^2+x+1=9\left(x-1\right)\Rightarrow x^2-8x+10=0\)

\(\Delta=\left(-8\right)^2-4.10=24\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{8-\sqrt{24}}{2}=4-\sqrt{6}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{8+\sqrt{24}}{2}=4+\sqrt{6}\end{matrix}\right.\)

\(TH_2:2b=a\Rightarrow2\sqrt{x^2+x+1}=\sqrt{x-1}\)

\(\Rightarrow4\left(x^2+x+1\right)=x-1\Rightarrow4x^2+3x+5=0\)

mà \(4x^2+3x+5=\left(2x\right)^2+2.2x.\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2+\dfrac{71}{16}=\left(2x+\dfrac{3}{4}\right)^2+\dfrac{71}{16}>0\)

\(\Rightarrow\) loại 

Vậy pt có tập nghiệm \(S=\left\{4+\sqrt{6};4-\sqrt{6}\right\}\)

\(5^{x+1}+5^{x-1}=130\)

\(5^x\cdot5^1+5^x\div5^1=130\)

\(5^x\cdot5^1+5^x\cdot\dfrac{1}{5}=130\)

\(5^x\cdot\left(5+\dfrac{1}{5}\right)=130\)

\(5^x\cdot\dfrac{26}{5}=130\)

\(5^x=130\div\dfrac{26}{5}\)

\(5^x=130\cdot\dfrac{5}{26}\)

\(5^x=25\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)

Mọi người còn câu trả lời nào khác không cứ trả lời đi mik tick cho

5^x+1.5^x-1=130

5^x-1.5²+5^x-1.1=130

5^x-1.(5²+1)=130

5^x-1.(25+1)=130

5^x-1.26=130

5^x-1=130:26

5^x-1=5¹

^{=> x-1=1}

^x=1+1

^x=2

^{Tick cho mình nha}