Tìm giá trị của x :
3(x+2)-2(x+1)= 26
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ P=\dfrac{x+16}{\sqrt{x}+3}\\ b,P=4\Leftrightarrow\dfrac{x+16}{\sqrt{x}+3}=4\\ \Leftrightarrow x+16=4\sqrt{x}+12\\ \Leftrightarrow x-4\sqrt{x}+4=0\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\\ \Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
\(c,P=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\\ P=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}-6=2\cdot5-6=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\Leftrightarrow\sqrt{x}+3=5\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,x=3-2\sqrt{2}\Leftrightarrow\sqrt{x}=\sqrt{2}-1\\ \Leftrightarrow P=\dfrac{3-2\sqrt{2}+16}{\sqrt{2}-1+3}=\dfrac{19-2\sqrt{2}}{\sqrt{2}+2}\\ P=\dfrac{\left(19-2\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2}=\dfrac{42-23\sqrt{2}}{2}\)
3(x-2)+2(x-1)=26
3x-6+2x-2=26
3x+2x-6-2=26
5x-8=26
5x=26+8
5x=34
x=34:5
x=6,8
Ta có:
3( x - 2 ) + 2 ( x + 1 ) = 26
3 x - 6 + 2 x + 2 = 26
3x + 2x = 26 + 6 - 2
5x = 30
x = 30 :5
x = 6
^^ Học tốt!
=> 3x+6-2x-2 = 26
=> x+4 = 26
=> x = 26-4 = 22
Vậy x=22
Tk mk nha
\(\Leftrightarrow3x+6-2x-2=26\)
\(\Leftrightarrow3x-2x=26-6+2\)
\(\Leftrightarrow1x=22\)
a) Để phân số \(\dfrac{26}{x+3}\) nguyên thì \(26⋮x+3\)
\(\Leftrightarrow x+3\in\left\{1;-1;2;-2;13;-13;26;-26\right\}\)
hay \(x\in\left\{-2-4;-1;-5;10;-16;23;-29\right\}\)
b) Để phân số \(\dfrac{x+6}{x+1}\) nguyên thì \(x+6⋮x+1\)
\(\Leftrightarrow5⋮x+1\)
\(\Leftrightarrow x+1\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{0;-2;4;-6\right\}\)
c) Để phân số \(\dfrac{x-2}{x+3}\) nguyên thì \(x-2⋮x+3\)
\(\Leftrightarrow-5⋮x+3\)
\(\Leftrightarrow x+3\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-2;-4;2;-8\right\}\)
d) Để phân số \(\dfrac{2x+1}{x-3}\) nguyên thì \(2x+1⋮x-3\)
\(\Leftrightarrow7⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{4;2;10;-4\right\}\)
a.17-(-x(-x(-x)))=16
17+x-x+x=16
17+x=16
x=16-17
x=-1
b.26-|x+19|=-13
|x+19|=26--13
|x+19|=26+13
|x+19|=39
\(\Rightarrow\)\(\orbr{\begin{cases}x+19=39\\x+19=-39\end{cases}}\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=20\\x=-58\end{cases}}\)
3(x + 2) - 2(x + 1) = 26
3x + 6 - 2x + 2 = 26
(3x - 2x) + (6 + 2) = 26
x + 8 = 26
x = 26 - 8
x = 18