a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)

b. \(\left(2x+1\right)+\left(4x+3\right)+\left(6x+5\right)+...+\left(100x+99\right)=7600\)

\(\rightarrow\left(2x+4x+6x+...+100x\right)+\left(1+3+5+...+99\right)=7600\)

\(\rightarrow\frac{\left(2x+100x\right).50}{2}+\frac{\left(1+99\right).50}{2}=7600\)

\(\rightarrow51x.50+50.50=7600\)

\(\rightarrow51x.50+2500=7600\)

\(\rightarrow51x.50=7600-2500\)

\(\rightarrow51x.50=5100\)

\(\rightarrow50x=100\)

\(\rightarrow x=\frac{100}{50}=2\)

Vậy x = 2

⇔5-x=0;6+6x=0;2x-4=0

TH₁: 5-x=0           TH₂:6+6x=0              TH₃:2x-4=0

   ⇔x=5                ⇔x=-1                        ⇔x=2

       Vậy x∈{5;-1;2}

* Trả lời:

\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)

\(\Leftrightarrow-3+6x-4-12x=-5x+5\)

\(\Leftrightarrow6x-12x+5x=3+4+5\)

\(\Leftrightarrow x=12\)

\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)

\(\Leftrightarrow6x-15-6+24x=-3x+7\)

\(\Leftrightarrow6x+24x+3x=15+6+7\)

\(\Leftrightarrow33x=28\)

\(\Leftrightarrow x=\dfrac{28}{33}\)

\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)

\(\Leftrightarrow1-3x-6x+12=-4x-5\)

\(\Leftrightarrow-3x-6x+4x=-1-12-5\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)

\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)

\(\Leftrightarrow-x-5x=-7\)

\(\Leftrightarrow-6x=-7\)

\(\Leftrightarrow x=\dfrac{7}{6}\)

\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)

\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)

\(\Leftrightarrow-15x+3x=4\)

\(\Leftrightarrow-12x=4\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

a) (12x-5)(4x-1)+(3x-7)(1-16x)

= (48x^2 - 12x - 20x + 5) + (3x - 48x^2 - 7 + 112x)

= 48x^2 - 12x - 20x + 5 +3x - 48x^2 -7 + 112x

= 83x-2

những phần sau bạn cứ làm tương tự theo cách nhân đa thức với đa thức và phá ngiawcj là ra nha :0))

sao ko làm hết ra

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

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

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

a) \(\left(x-7\right)^2-x.\left(x+3\right)=0\)

⇒ \(x^2-2.7.x+7^2-\left(x^2+3x\right)=0\)

⇒ \(x^2-14x+49-x^2-3x=0\)

⇒ \(-17x+49=0\)

⇒ \(-17x=0-49\)

⇒ \(-17x=-49\)

⇒ \(x=\left(-49\right):\left(-17\right)\)

⇒ \(x=\frac{49}{17}\)

Vậy \(x=\frac{49}{17}.\)

Chúc bạn học tốt!

mơn bn nhìu