(7x−12)+(8x−10)+(3x−14)=4.
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20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
a: \(\dfrac{x^2-5x+6}{x^2+7x+12}:\dfrac{x^2-4x+4}{x^2+3x}\)
\(=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+3\right)\left(x+4\right)}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)
\(=\dfrac{x\left(x-3\right)}{\left(x-2\right)\left(x+4\right)}\)
b: \(\dfrac{x^2+2x-3}{x^2+3x-10}:\dfrac{x^2+7x+12}{x^2-9x+14}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x+5\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x-7\right)}{\left(x+3\right)\left(x+4\right)}\)
\(=\dfrac{\left(x-1\right)\left(x-7\right)}{\left(x+5\right)\left(x+4\right)}\)
Mấy cái này chuyển vế đổi dấu là xong í mà :3
1,
16-8x=0
=>16=8x
=>x=16/8=2
2,
7x+14=0
=>7x=-14
=>x=-2
3,
5-2x=0
=>5=2x
=>x=5/2
Mk làm 3 cau làm mẫu thôi
Lúc đăng đừng đăng như v :>
chi ra khỏi ngt nản
từ câu 1 đến câu 8 cs thể làm rất dễ,bn tham khảo bài của bn muwaa r làm những câu cn lại
\(D=\dfrac{x-2}{x+2}.\left(\dfrac{5x+10}{7x-14}+\dfrac{x-2}{3x-6}\right)+\dfrac{3x^2-12}{2x^2-8x+8}\)
\(D=\dfrac{x-2}{x+2}.\left(\dfrac{5\left(x+2\right)}{7\left(x-2\right)}+\dfrac{x-2}{3\left(x-2\right)}\right)+\dfrac{3\left(x^2-4\right)}{2\left(x^2-4x+4\right)}\)
\(D=\dfrac{x-2}{x+2}.\dfrac{5\left(x+2\right)}{7\left(x-2\right)}+\dfrac{x-2}{3\left(x-2\right)}.\dfrac{x-2}{x+2}+\dfrac{3\left(x^2-4\right)}{2\left(x^2-4x+4\right)}\)
\(D=\dfrac{5}{7}+\dfrac{x-2}{2\left(x+2\right)}+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x-2\right)^2}\)
\(D=\dfrac{5}{7}+\dfrac{x-2}{2\left(x+2\right)}+\dfrac{3\left(x+2\right)}{2\left(x-2\right)}\)
\(D=\dfrac{5}{7}-\dfrac{-\left(x-2\right)}{2\left(x-2\right)}+\dfrac{3\left(x+2\right)}{2\left(x-2\right)}\)
\(D=\dfrac{5}{7}-\dfrac{-\left(x-2\right)+3x+2}{2\left(x-2\right)}\)
\(D=\dfrac{5}{7}-\dfrac{2x+4}{2\left(x-2\right)}\)
\(D=\dfrac{5}{7}+\dfrac{2\left(x-2\right)}{2\left(x-2\right)}=\dfrac{5}{7}+\dfrac{x-2}{x-2}\)
\(D=\dfrac{5}{7}+1=\dfrac{12}{7}\)
Vậy \(D=\dfrac{12}{7}\)
\(B=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7x-14}+\dfrac{x-2}{3x-6}\right)+\dfrac{3\left(x^2-4\right)}{2x^2-8x+8}\)
\(=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7\left(x-2\right)}+\dfrac{x-2}{3\left(x-2\right)}\right)+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x^2-4x+4\right)}\)
\(=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7\left(x-2\right)}+\dfrac{1}{3}\right)+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x-2\right)^2}\)
\(=\dfrac{x-2}{x+2}\cdot\dfrac{3\left(5x+10\right)+7\left(x-2\right)}{21\left(x-2\right)}+\dfrac{3\left(x+2\right)}{2\left(x-2\right)}\)
\(=\dfrac{1}{x+2}\cdot\dfrac{15x+30+7x-14}{21}+\dfrac{3x+6}{2\left(x-2\right)}\)
\(=\dfrac{22x+16}{21\left(x+2\right)}+\dfrac{3x+6}{2\left(x-2\right)}\)
\(=\dfrac{2\left(x-2\right)\left(22x+16\right)+21\left(x+2\right)\left(3x+6\right)}{42\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{\left(2x-4\right)\left(22x+16\right)+\left(21x+42\right)\left(3x+6\right)}{42\left(x^2-4\right)}\)
\(=\dfrac{44x^2+32x-88x-64+63x^2+126x+126x+252}{42x^2-168}\)
\(=\dfrac{107x^2+196x+188}{42x^2-168}\)
Lời giải:
$(7x-12)+(8x-10)+(3x-14)=4$
$(7x+8x+3x)-(12+10+14)=4$
$18x-36=4$
$18x=40$
$x=\frac{20}{9}$
\(7x-12+8x-10+3x-14-4=0\)
\(18x-40=0\)
\(x=\dfrac{40}{18}=\dfrac{20}{9}\)