1) \(M=9x^2-6x+6=\left(9x^2-6x+1\right)+5=\left(3x-1\right)^2+5\ge5\)

\(minM=5\Leftrightarrow x=\dfrac{1}{3}\)

2) \(M=5-2x-x^2=-\left(x^2+2x+1\right)+6=-\left(x+1\right)^2+6\le6\)

\(maxM=6\Leftrightarrow x=-1\)

3) \(N=5+6x-9x^2=-\left(9x^2-6x+1\right)+6=-\left(3x-1\right)^2+6\le6\)

\(maxN=6\Leftrightarrow x=\dfrac{1}{3}\)

Vì 1⁵= 1⁶ nên x - 5 = 1

x= 1+5

x=6

\(\left(x-5\right)^5=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^5=0\)

\(\Rightarrow\left(x-5\right)^5\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

\(\left(x-5\right)^5=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^5=0\)

\(\Rightarrow\left(x-5\right)^5.\left(x-5-1\right)=0\)

\(\Rightarrow\left(x-5\right)^5.\left(x-6\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-5\right)^5=0\\x-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

=41/10

x - 1/2 = 18/5

x = 18/5 + 1/2

x = 41/10

\(x-\dfrac{1}{2}=\dfrac{3}{5}\times6\)

\(x-\dfrac{1}{2}=\dfrac{18}{5}\)

\(x=\dfrac{18}{5}+\dfrac{1}{2}\)

\(x=\dfrac{41}{10}\)

Bài 4:

a: =>7/x-5=2

=>x-5=7/2

=>x=17/2

b: =>1-2x=-5

=>2x=6

=>x=3

c: =>2x-3=5 hoặc 2x-3=-5

=>2x=8 hoặc 2x=-2

=>x=-1 hoặc x=4

d: =>2(x+1)^2+17=21

=>2(x+1)^2=4

=>(x+1)^2=2

=>\(x+1=\pm\sqrt{2}\)

=>\(x=\pm\sqrt{2}-1\)

\(2^x-26=6\\ \Leftrightarrow2^x=32\\ \Leftrightarrow2^x=2^5\\ \Leftrightarrow x=5\)

\(2^x-26=6\)

⇒ \(2^x=6+26=32\)

⇒ \(2^x=32=2^5\)

⇒ \(x=5\)

\(2^x-26=6\)

\(2^x=6+26=32\)

\(2^x=2^5\)

=> x = 5

Vậy x = 5

\(\sqrt{x+1}+\sqrt{4x+4}=6\)

\(\Leftrightarrow\sqrt{x+1}+4\sqrt{x+1}=6\)

\(\Leftrightarrow5\sqrt{x+1}=6\)

\(\Leftrightarrow\sqrt{x+1}=\dfrac{6}{5}\)

\(\Leftrightarrow x+1=\left(\dfrac{6}{5}\right)^2\)

\(\Leftrightarrow x+1=\dfrac{36}{25}\)

`<=> x= 11/25`

** Bổ sung điều kiện $x,y$ là các số nguyên. 

$x+5y+xy=6$

$(x+xy)+5y=6$

$x(1+y)+5(y+1)=11$

$(y+1)(x+5)=11$
Vì $x,y$ nguyên nên $x+5, y+1$ cũng nguyên. Ta xét các TH sau:

TH1: $x+5=1, y+1=11\Rightarrow x=-4; y=10$
TH2: $x+5=11, y+1=1\Rightarrow x=6; y=0$

TH3: $x+5=-1; y+1=-11\Rightarrow x=-6; y=-12$

TH4: $x+5=-11; y+1=-1\Rightarrow x=-16; y=-2$

giúp mình với ạ mình cần gấp

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

⇒\(\dfrac{y-x}{5-2}=\dfrac{6}{3}=2\)

\(\dfrac{x}{2}=2\Rightarrow x=4\)

\(\dfrac{y}{5}=2\Rightarrow y=10\)

\(\dfrac{z}{10}=2\Rightarrow z=20\)

b) Ta có: \(\dfrac{x}{8}=\dfrac{2y}{6}=\dfrac{z}{7}\)

\(\dfrac{x-2y+z}{8-6+7}=\dfrac{18}{9}=2\)

\(\dfrac{x}{8}=2\Rightarrow x=16\)

\(\dfrac{y}{3}=2\Rightarrow y=6\)

\(\dfrac{z}{7}=2\Rightarrow z=14\)

Để (d1) cắt (d2) tại một điểm trên trục tung thì

\(\left\{{}\begin{matrix}m-2\ne-2\\m^2+5m+6=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m\ne0\\m^2+5m=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\m\left(m+5\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m\ne0\\\left[{}\begin{matrix}m=0\\m+5=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow m+5=0\)

=>m=-5