a.

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x-2}=b\end{matrix}\right.\) ta được:

\(2a^2-b^2=ab\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\8a^3=-b^3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(vô-nghiệm\right)\\8\left(x+2\right)=-\left(x-2\right)\end{matrix}\right.\)

\(\Leftrightarrow x=-\dfrac{14}{9}\)

b.

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\)

\(\Rightarrow a^2+4b^2=5ab\)

\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\a^3=64b^3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}65+x=65-x\\65+x=64\left(65-x\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

c.

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x+1}=b\end{matrix}\right.\)

\(\Rightarrow a+b=1+ab\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=1\\b^3=1\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

`a,`\(2^x -15= 2^4+1\)

`-> 2^x-15=17`

`-> 2^x=17+15`

`-> 2^x=32`

`-> 2^x=2^5`

`-> x=5`

`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?

`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)

`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)

`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)

`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)

Mà `1/65+1/64-1/63-1/62 \ne 0`

`-> x+66=0`

`-> x=-66`

a: =>2^x=2^4+16=32

=>x=5

b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)

=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)

=>x+66=0

=>x=-66

b, ĐKXĐ: \(x\ge\frac{5}{2}\)

\(pt\Leftrightarrow\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}+1\right)^2}=4\)

\(\Leftrightarrow\sqrt{2x-5}=3\)

\(\Leftrightarrow x=7\left(tm\right)\)

a, ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{x-5+4\sqrt{x-5}+4}+\sqrt{x-5+8\sqrt{x-5}+16}=0\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-5}+2\right)^2}+\sqrt{\left(\sqrt{x-5}+4\right)^2}=0\)

\(\Leftrightarrow2\sqrt{x-5}+6=0\)

\(\Leftrightarrow\sqrt{x-5}=-3\)

Phương trình vô nghiệm

ai nhanh nhất mình k cho

Bài 1:

\(A=\frac{1}{\left(1+2\right)}+\frac{1}{\left(1+2+3\right)}+\frac{1}{\left(1+2+3+4\right)}\)\(+\frac{1}{\left(1+2+3+4+5\right)}+...+\)\(\frac{1}{\left(1+2+3+...+10\right)}\)

\(A=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{55}\)

\(A=2\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)

\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(A=\frac{9}{11}\)

Bài 2 :

2)  Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15

= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15

Mẫu số =  11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19  + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17 lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17  

=  (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x  17 = (1+27+125+729) x 13 x 15 x 17 

\(\Rightarrow A< \frac{\left(1+27+125+729\right)\times11\times13\times15}{\left(1+27+125+729\right)\times13\times15\times17}\)

\(=\frac{11}{17}\)

\(=\frac{1111}{1717}=B\)

Vậy \(A=B\)

mình đâu biết

( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 65

( x + x + x + x + x ) + ( 1 + 2 + 3 + 4 = 5 ) = 65

x * 5 +15 = 65

x * 5 = 65 - 15

x * 5 = 50

x = 50 : 5

x = 10

( X + 1 ) + ( X + 2 ) + ( X + 3 ) + ( X + 4 ) + ( X + 5 ) = 65

X x 5 + ( 1 + 2 + 3 + 4 + 5 ) = 65

X x 5 + 15 = 65

X x 5 = 65 - 15

X x 5 = 50

      X = 50 : 5

      X = 10

x=10

ung ho nha minh chuan 100% nha

(x+1) + ( x+2) + (x+3) + (x+4) + ( x+5)= 65

(x+x+x+x+x)+(1+2+3+4+5)=65

x*5+15=65

x*5=65-15

x*5=50

x=10

<=>(x+x+x+x+x)+(1+2+3+4+5)=65

=>5x+15=65

=>5x=65-15

=>5x=50

=>x=50:5

=>x=10

=>x+1+x+2+x+3+x+4+x+5=65

=>(x+x+x+x+x)+(1+2+3+4+5)=65

=>5x+15=65

=>5x=50 tức là 65-15

=>x=10

a) -65 .( 87 - 17 ) -87 .( 17 - 65 )

= ( - 65 ) . 87 + 17 - 87 . 17 + 65

= { ( - 65 ) + 65 } . 87 + 17 

= 0 . 87 + 17

= 17

b) -215 . [ 14 + ( -236 ) ] + 215 . ( 14 - 236 )

= -215 . 14 + ( - 236 ) + 215 . 14 - 236

= [ ( - 215 ) + 215 ] . 14 + { ( - 236 - 236 } 

= 0 . 14 + 0 

= 0 

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+\left(x+5\right)=65\)

<=>  \(5x+15=65\)

<=> \(5x=50\)

<=> \(x=10\)

Vậy...

p/s: chúc bạn học tốt