Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. \(\dfrac{6x+5}{2}-\dfrac{10x+3}{4}=2x+\dfrac{2x+1}{2}\)
\(\Leftrightarrow2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
\(\Leftrightarrow12x+10-10x-3=8x+4x+2\)
\(\Leftrightarrow12x-10x-8x-4x=2-10+3\)
\(\Leftrightarrow-10x=-5\Leftrightarrow x=\dfrac{1}{2}\)
b. \(\left(x+1\right)^3-\left(x-1\right)^3=6\left(x^2+x+1\right)\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+6x+6\)
\(\Leftrightarrow6x^2+2=6x^2+6x+6\)
\(\Leftrightarrow6x^2-6x^2-6x=6-2\Leftrightarrow-6x=4\)
\(\Leftrightarrow x=\dfrac{-2}{3}\)
c. \(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\left(\dfrac{x+2}{13}+1\right)+\left(\dfrac{2x+45}{15}-1\right)=\left(\dfrac{3x+8}{37}+1\right)+\left(\dfrac{4x+69}{9}-1\right)\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}-\dfrac{3\left(x+15\right)}{37}-\dfrac{4\left(x+15\right)}{9}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
Vì \(\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)>0\)
\(\Leftrightarrow x+15=0\Leftrightarrow x=-15\)
b) \(\dfrac{5\left(4x-1\right)}{15}-\dfrac{2-x}{15}-\dfrac{3\left(10x-3\right)}{15}\le0\)
\(\Leftrightarrow\dfrac{20x-5-2+x-30x+9}{15}\le0\)
\(\Rightarrow-9x+2\le0\)
\(\Leftrightarrow-9x\le-2\)
\(\Rightarrow-9x.\dfrac{-1}{9}\ge-2.\dfrac{-1}{9}\)
\(\Leftrightarrow x\ge\dfrac{2}{9}\)
câu a ,không hiểu đề
a)
\(\left(x-3\right)^2< x^2-5x+4\)
\(\Leftrightarrow x^2-6x+9< x^2-5x+4\)
\(\Leftrightarrow x^2-x^2-6x+5x< 4-9\)
\(\Leftrightarrow-x>-5\)
\(\Leftrightarrow x>5\)
Vây...
b)
\(\left(x-3\right)\left(x+3\right)\le\left(x+2\right)^2+3\)
\(\Leftrightarrow x^2-9\le x^2+4x+9\)
\(\Leftrightarrow x^2-x^2-4x\le9+9\)
\(\Leftrightarrow-4x\le18\)
\(\Leftrightarrow x\ge-4,5\)
Vậy....
Bạn tự biểu diễn trên trục số ha!
c)
\(\dfrac{4x-7}{3}>\dfrac{7-x}{5}\)
\(\Leftrightarrow15.\dfrac{4x-5}{3}< 15.\dfrac{7-x}{5}\)
\(\Leftrightarrow5.\left(4x-5\right)< 3.\left(7-x\right)\)
\(\Leftrightarrow24x-25< 21-3x\)
\(\Leftrightarrow20x+3x< 21+25\)
\(\Leftrightarrow23x< 46\)
\(\Leftrightarrow x< 2\)
Vậy...
d)
\(\dfrac{x+2}{x-3}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2< 0\\x-3< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-2\\x< 3\end{matrix}\right.\)
Vậy...
1.
|x-9|=2x+5
x<9; x-9=-2x-5
3x=4=>x=4/3(n)
x≥9; x-9=2x+5=> x=-14(l)
2.a
A=2x-5≥0<=>2x≥5; x≥5/2
1. a) / x - 9 / = 2x + 5
Do : / x - 9 / ≥ 0 ∀x
⇒2x + 5 ≥ 0
⇔ x ≥ \(\dfrac{-5}{2}\)
Bình phương cả hai vế của phương trình , ta được :
( x - 9)2 = ( 2x + 5)2
⇔ ( x - 9)2 - ( 2x + 5)2 = 0
⇔ ( x - 9 - 2x - 5)( x - 9 + 2x + 5) = 0
⇔ ( - x - 14)( 3x - 4) = 0
⇔ x = - 14 ( KTM) hoặc : x = \(\dfrac{4}{3}\) ( TM)
KL....
b) Mạn phép làm luôn , ko chép lại đề :
\(\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-5}{\left(x-3\right)\left(x+3\right)}\) ( x # 3 ; x # - 3)
⇔ 5x + 15 + 4x - 12 = x - 5
⇔ 9x + 3 = x - 5
⇔ 8x = - 8
⇔ x = -1 ( TM)
KL....