a: \(B=\dfrac{10x}{\left(x+4\right)\left(x-1\right)}-\dfrac{2x-3}{x+4}-\dfrac{x+1}{x-1}\)

\(=\dfrac{10x-\left(2x^2-2x-3x+3\right)-\left(x^2+5x+4\right)}{\left(x+4\right)\left(x-1\right)}\)

\(=\dfrac{10x-2x^2+5x-3-x^2-5x-4}{\left(x+4\right)\left(x-1\right)}\)

\(=\dfrac{-3x^2+10x-7}{\left(x+4\right)\left(x-1\right)}\)

\(=\dfrac{-\left(3x^2-10x+7\right)}{\left(x-1\right)\left(x+4\right)}=-\dfrac{\left(x-1\right)\left(3x-7\right)}{\left(x-1\right)\left(x+4\right)}\)

\(=\dfrac{-3x+7}{x+4}\)

b: \(B+3=\dfrac{-3x+7+3x+12}{x+4}=\dfrac{19}{x+4}>0\)

=>B>-3

a/x^4 lớn hơn hoặc = 0 

x^2 lớn hơn hoặc = 0

2 > 0

=> x^4+x^2+2 >0 => bieu thức luôn dương

b/ (x+3)(x-11)+2003 <=> x^2 -8x -33 +2003 <=> x^2 -8x +1970 <=> x^2-8x+16+1954 <=> (x-4)^2+1954 

ta có : (x-4)^2 lớn hơn hoặc = 0

           1954 >0

=> (x-4)^2+1954>0 => bt luôn dương

Bài 1 trước nha . chúc bạn học tốt . Ủng hộ nha

\(=>-9\left(x^2-\frac{4}{3}x+\frac{5}{3}\right)=>-9\left(x^2-2.\frac{2}{3}x+\frac{4}{9}+\frac{11}{9}\right)=>-9\left(x-\frac{2}{3}\right)^2-11\)

Ta có \(\left(x-\frac{2}{3}\right)^2\ge0=>-9\left(x-\frac{2}{3}\right)^2\le0,-11< 0\)

\(-9\left(x-\frac{2}{3}\right)^2-11\le0\)=> bt luôn âm

\(=>-5-x^2-x+2=>-x^2-x-3=>-x^2-2.\frac{1}{2}x+\frac{1}{4}-\frac{13}{4}\)\(=>-\left(x-\frac{1}{2}\right)^2-\frac{13}{4}\)

Ta có \(\left(x-\frac{1}{2}\right)^2\ge0=>=>-\left(x-\frac{1}{2}\right)^2\le0,-\frac{13}{4}< 0\)

\(=>-\left(x-\frac{1}{2}\right)^2-\frac{13}{4}< 0\)=> bt luôn âm

ùng hộ mình nha. cảm ơn

a) Đặt A(x)=0

\(\Leftrightarrow4x-1=0\)

\(\Leftrightarrow4x=1\)

hay \(x=\dfrac{1}{4}\)

b) Đặt B(x)=0

\(\Leftrightarrow2x^2-8=0\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

c) Đặt C(x)=0

\(\Leftrightarrow x^2+1=0\)

mà \(x^2+1\ge1>0\forall x\)

nên \(x\in\varnothing\)

a) \(2x\left(x-3\right)-x\left(2x+1\right)-3\left(x+5\right)=11\)

\(\Rightarrow2x^2-6x-2x^2-x-3x-15=11\)

\(\Rightarrow-10x=26\Rightarrow x=-2,6\)

Vậy ...........

b) \(x\left(x-1\right)-\left(x^2+3x-5\right)-2\left(x+3\right)=17\)

\(\Rightarrow x^2-x-x^2-3x+5-2x-6=17\)

\(\Rightarrow-6x=18\Rightarrow x=-3\)

c) \(5x\left(x-7\right)-\left(5x+1\right)x-\left(x+3\right)2=13\)

\(\Rightarrow5x^2-35x-5x^2-x-2x-6=13\)

\(\Rightarrow-38x=19\Rightarrow x=-\frac{1}{2}\)

d) \(\left(2x^2-3x+5\right)-2x\left(x-3\right)+\left(x-1\right)\left(-2\right)=10\)

\(\Rightarrow2x^2-3x+5-2x^2+6x-2x+2=10\)

\(\Rightarrow x=3\)

giúp mik với

Cảm ơn bạn Diễm Quỳnh nhiều nha.

a) 4.(18- 5x) - 12.( 3x-7) =15.(2x-16) - 6.(x+14)

72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

-20x - 36x - 30x + 6x = -240 - 84 - 72 -84 

-80x = -480

x= 6

4.(18 - 5x) - 12.(3x - 7) = 15.(2x - 16) - 6.(x+14)
4.18 - 4.5x - 12.3x + 12.7 = 15.2x - 15.16 - 6x - 6.14
72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
72 + 84 - 20x - 36x = 30x - 6x - 240 - 84
156 - 56x               = 24x - 324
156                        = 24x - 324 + 56x
156                       = 80x - 324
80x - 324               = 156
80x                       = 156 + 324
80x                       = 480
x                           = 480:80

x                           = 6
câu b giải tương tự  

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

5)

để \(\frac{5x-3}{x+1}\)là số nguyên

\(5x-3⋮x+1\)

\(x+1⋮x+1\)

\(\Rightarrow5\left(x+1\right)⋮x+1\)

\(5x-3-\left(5x-5\right)⋮x+1\)

\(-2⋮x+1\)

\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Vậy \(x\in\left\{0;-2;1;-3\right\}\)