b: \(3x^2-2x-1=0\)

=>\(3x^2-3x+x-1=0\)

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

=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

a: Bạn ghi lại đề đi bạn

a: \(\Leftrightarrow3x=-21\)

hay x=-7

a

-3x=33-12

3x= -21

x= -7

-x²+x+2-x-3x²-11x-10+4x²-2=0

-10x-10=0

x=-1

a) \(\dfrac{2x+5}{2x+1}=\dfrac{2x+1+4}{2x+1}=\dfrac{2x+1}{2x+1}+\dfrac{4}{2x+1}=1+\dfrac{4}{2x+1}\)  

Để \(\dfrac{2x+5}{2x+1}\in Z\) thì \(\dfrac{4}{2x+1}\in Z\) 

\(\Rightarrow4\) ⋮ \(2x+1\)

\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow2x\in\left\{0;-2;1;-3;3;-5\right\}\)

\(\Rightarrow x\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2};\dfrac{3}{2};-\dfrac{5}{2}\right\}\)

Mà x nguyên \(\Rightarrow\text{x}\in\left\{0;-1\right\}\) 

b) \(\dfrac{3x+5}{x+1}=\dfrac{3x+3+2}{x+1}=\dfrac{3\left(x+1\right)+2}{x+1}=\dfrac{3\left(x+1\right)}{x+1}+\dfrac{2}{x+1}=3+\dfrac{2}{x+1}\) 

Để \(\dfrac{3x+5}{x+1}\in Z\) thì \(\dfrac{2}{x+1}\in Z\) 

\(\Rightarrow2\) ⋮ \(x+1\)

\(\Rightarrow x+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

\(\Rightarrow x\in\left\{0;-2;1;-3\right\}\) 

c) \(\dfrac{3x+8}{x-1}=\dfrac{3x-3+11}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{11}{x-1}=3+\dfrac{11}{x-1}\)  

Để: \(\dfrac{3x+8}{x-1}\in Z\) thì \(\dfrac{11}{x-1}\in Z\)

\(\Rightarrow11\) ⋮ \(x-1\)

\(\Rightarrow x-1\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)

\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)

d) \(\dfrac{5x+12}{x-2}=\dfrac{5x-10+22}{x-2}=\dfrac{5\left(x-2\right)+22}{x-2}=\dfrac{5\left(x-2\right)}{x-2}+\dfrac{22}{x-2}=5+\dfrac{22}{x-2}\)

Để: \(\dfrac{5x+12}{x-2}\in Z\) thì \(\dfrac{22}{x-2}\in Z\)

\(\Rightarrow22\) ⋮ \(x-2\)

\(\Rightarrow x-2\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)

\(\Rightarrow x\in\left\{3;1;4;0;13;-9;24;-20\right\}\)

e) \(\dfrac{7x-12}{x+16}=\dfrac{7x+112-124}{x+16}=\dfrac{7\left(x+16\right)-124}{x+16}=\dfrac{7\left(x+16\right)}{x+16}-\dfrac{124}{x+16}=7-\dfrac{124}{x+16}\)

Để \(\dfrac{7x-12}{x+16}\in Z\) thì \(\dfrac{124}{x+16}\in Z\) 

\(\Rightarrow124\) ⋮ \(x+16\)

\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)

\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)

\(x-7+3x=5-3x+1+3x-\left(-3\right)\)

\(x-7+3x=5-3x+1+3x+3\)

\(x+3x-3x=7+5+1+3\)

\(x=16\)

a, 2 x ( 15 - 3x ) = 12

=>  15 - 3x  =  6

=>  3x  = 9

=>  x  =  3

b, 39 - 2 x ( 31 - 3x ) = 15

=>  2 x ( 31 - 3x ) =  24

=>  31 - 3x  = 12

=>  3x  = 19

=>  x  =  \(\frac{19}{3}\)

nếu mk sửa đề có sai thì nhờ bạn sửa đề lại giúp mk nha

\(2\left(15-3x\right)\)\(=12\)

\(15-3x=12:2=6\)

\(3x=15-6=9\)

\(x=9:3=3\)

\(b,39-2\left(31-3x\right)\)\(-15\)

Đến đây cho anh hỏi thế biểu thức này giá trị bằng bao nhiêu

a) Ta có: \(\left|x+5\right|\ge0\forall x\)

\(\Rightarrow\left|x+5\right|+2023\ge2023\forall x\)

\(\Rightarrow A\ge2023\forall x\)

Dấu \("="\) xảy ra khi: \(x+5=0\Leftrightarrow x=-5\)

Vậy \(Min_A=2023\) khi \(x=-5\).

b) Ta có: \(\left\{{}\begin{matrix}\left|2x+6\right|\ge0\forall x\\\left|y+3x\right|\ge0\forall x,y\end{matrix}\right.\)

\(\Rightarrow\left|2x+6\right|+\left|y+3x\right|\ge0\forall x,y\)

\(\Rightarrow\left|2x+6\right|+\left|y+3x\right|+25\ge25\forall x,y\)

\(\Rightarrow B\ge25\forall x,y\)

Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}2x+6=0\\y+3x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-6\\y=-3x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-6:2=-3\\y=-3\cdot\left(-3\right)=9\end{matrix}\right.\)

Vậy \(Min_B=25\) khi \(x=-3;y=9\).

c) Ta có: \(\left\{{}\begin{matrix}\left|12-3x\right|\ge0\forall x\\\left|-y-4x\right|\ge0\forall x,y\end{matrix}\right.\)

\(\Rightarrow\left|12-3x\right|+\left|-y-4x\right|\ge0\forall x,y\)

\(\Rightarrow\left|12-3x\right|+\left|-y-4x\right|-12\ge-12\forall x,y\)

\(\Rightarrow C\ge-12\forall x,y\)

Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}12-3x=0\\-y-4x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=12\\y=-4x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=12:3=4\\y=-4\cdot4=-16\end{matrix}\right.\)

Vậy \(Min_C=-12\) khi \(x=4;y=-16\).

\(\mathit{Toru}\)

a: \(T=\dfrac{3}{2}x^4-x^3+3x^2-\dfrac{1}{2}x+6+x^4+\dfrac{2}{3}x^3-2x^2-4x+1\)

\(=\dfrac{5}{2}x^4-\dfrac{1}{3}x^3+x^2-\dfrac{9}{2}x+7\)

b: \(T\left(2\right)=\dfrac{5}{2}\cdot16-\dfrac{1}{3}\cdot8+4-\dfrac{9}{2}\cdot2+7=\dfrac{118}{3}\)

a) (3x – 5)2 – (x +1 )2 = (3x – 5 – x – 1)(3x – 5 + x + 1)

= (2x – 6)(4x – 4) = 8(x – 1)(x – 3)

Vậy (x – 1)(x – 3) = 0 ⇒ x - 1 = 0 hoặc x - 3 = 0

⇒ x = 1hoặc x = 3

a: =>2x+5=9

=>2x=4

hay x=2

b: =>3x-2x+12=12

=>x=0