Bài 3: Phương trình đưa được về dạng ax + b = 0

Gọi lượng công việc mà 2 người đó làm là x;

- Năng suất lao động của người 1 là a ; năng suất lao động của người 2 là b; Theo bài ra ta có:

\(x=\left(a+b\right).15=>x=15a+15b\) và :

\(x=\left(a+b\right).8+21b=8a+29b\)

hay \(=>x=15a+15b-7a+14b\)

\(=>x=x-7\left(a-2b\right)=>7\left(a-2b\right)=0\)

Từ đó ta có : \(a=2b\) ( năng suất người 1 gấp đôi người 2) ;

Thay a= 2b vào x thì ta có:

\(=>x=15.2b+15.b=45b\)

\(=>x=15.a+\dfrac{15}{2}b=22,5a\)

Vậy người thứ nhất làm 1 mình hết 22,5 giờ ; người thứ hai làm một mình hết 45 giờ;

CHÚC BẠN HỌC TỐT......

\(x\left(x+2\right)\left(x^2+2x+5\right)=6\)

\(\Leftrightarrow\left(x^2+2x\right)\left(x^2+2x+5\right)-6=0\)

Đặt \(t=x^2+2x\) thì ta có:

\(t\left(t+5\right)-6=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t-1=0\\t+6=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+2x=1\\x^2+2x=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x^2+2x-1=0\\x^2+2x+6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+2x+1-2=0\\x^2+2x+1+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2-2=0\\\left(x+1\right)^2+5>0\end{matrix}\right.\)

\(\Rightarrow\left(x+1\right)^2=2\Rightarrow x+1=\pm\sqrt{2}\Rightarrow x=\pm\sqrt{2}-1\)

\((6x+1)^2-2(x+1)^3+2(x-1)(x^2+x+1)=1\)

\(\Leftrightarrow36x^2+12x+1-2\left(x^3+3x^2+3x+1\right)+2\left(x^3-1\right)=1\)

\(\Leftrightarrow36x^2+12x+1-2x^3-6x^2-6x-2+2x^3-2=1\)

\(\Leftrightarrow30x^2+6x-4=0\)\(\Leftrightarrow2\left(15x^2+3x-2\right)=0\)

\(\Leftrightarrow15x^2+3x-2=0\)\(\Leftrightarrow15\left(x+\dfrac{1}{10}\right)^2-\dfrac{43}{20}=0\)

\(\Leftrightarrow15\left(x+\dfrac{1}{10}\right)^2=\dfrac{43}{20}\Leftrightarrow x=\pm\dfrac{\sqrt{129}}{30}-\dfrac{1}{10}\)

\(\dfrac{\left(7x+1\right)\left(x-2\right)}{10}+\dfrac{2}{5}=\dfrac{\left(x-2\right)^2}{5}+\dfrac{\left(x-1\right)\left(x-3\right)}{2}\)

\(\Leftrightarrow\dfrac{7x^2-14x+x-2+4}{10}=\dfrac{2\left(x-2\right)^2+5\left(x^2-3x-x+3\right)}{10}\) \(\Leftrightarrow7x^2-13x+2=2x^2-8x+8+5x^2-15x-5x+15\)

\(\Leftrightarrow7x^2-13x+2=7x^2-28x+23\)

\(\Leftrightarrow7x^2-13x+2-7x^2+28x-23=0\)

\(\Leftrightarrow15x-21=0\)

\(\Leftrightarrow15x=21\)

\(\Leftrightarrow x=\dfrac{7}{5}\)

Vậy \(x=\dfrac{7}{5}\).

Bạn chuyển 2 vế lại vs nhau : vế có x và vế có phân số rồi tìm :))

Chúc bạn học tốt :))

1a

\(\dfrac{1}{5-2x}+\dfrac{2x-2}{6x-15}\\ =\dfrac{3}{15-6x}-\dfrac{2x-2}{15-6x}\\ =\dfrac{3-2x+2}{15-6x}=\dfrac{5-2x}{15-6x}=\dfrac{1}{3}\)

b.

\(\dfrac{2x+6}{x-3}:\dfrac{x^2+6x+9}{4x^2-36}\\ =\dfrac{2x+6}{x-3}.\dfrac{4\left(x-3\right)\left(x+3\right)}{\left(x+3\right)^2}\\ =8\)

\(a,\dfrac{3\left(5x-2\right)}{4}-2=\dfrac{7x}{3}-5\left(x-7\right)\)

\(\Leftrightarrow\dfrac{15x-6-8}{4}=\dfrac{7x-15\left(x-7\right)}{3}\)

\(\Leftrightarrow\dfrac{15x-14}{4}=\dfrac{7x-15x+105}{3}\)

\(\Leftrightarrow\dfrac{45x-42}{12}=\dfrac{-32x+420}{12}\)

\(\Leftrightarrow45x+32x=420+42\)

\(\Leftrightarrow77x=462\)

\(\Leftrightarrow x=6\)

\(b,\dfrac{x+5}{2}+\dfrac{3-2x}{4}=x-\dfrac{7+x}{6}\)

\(\Leftrightarrow\dfrac{2x+10+3-2x}{4}=\dfrac{6x-7-x}{6}\)

\(\Leftrightarrow\dfrac{13}{4}=\dfrac{5x-7}{6}\)

\(\Leftrightarrow2\left(5x-7\right)=3.13\)

\(\Leftrightarrow10x-14=39\)

\(\Leftrightarrow10x=53\)

\(\Leftrightarrow x=5,3\)

\(c,\dfrac{x-3}{11}+\dfrac{x+1}{3}=\dfrac{x+7}{9}-1\)

\(\Leftrightarrow\dfrac{3x-9+11x+11}{33}=\dfrac{x+7-9}{9}\)

\(\Leftrightarrow\dfrac{14x+2}{33}=\dfrac{x-2}{9}\)

\(\Leftrightarrow33\left(x-2\right)=9\left(14x+2\right)\)

\(\Leftrightarrow33x-66=126x+18\)

\(\Leftrightarrow-93x=84\)

\(\Leftrightarrow x=-\dfrac{28}{31}\)

\(d,\dfrac{3x-0,4}{2}+\dfrac{1,5-2x}{3}=\dfrac{x+0,5}{5}\)

\(\Leftrightarrow\dfrac{3\left(3x-0,4\right)+2\left(1,5-2x\right)}{6}=\dfrac{x+0,5}{5}\)

\(\Leftrightarrow\dfrac{9x-1,2+3-4x}{6}=\dfrac{x+0,5}{5}\)

\(\Leftrightarrow\dfrac{5x+1,8}{6}=\dfrac{x+0,5}{5}\)

\(\Leftrightarrow5\left(5x+1,8\right)=6\left(x+0,5\right)\)

\(\Leftrightarrow25x+9=6x+3\)

\(\Leftrightarrow19x=-6\)

\(\Leftrightarrow x=-\dfrac{6}{19}\)

\(\Leftrightarrow77x=378\)

\(\Leftrightarrow x=\dfrac{54}{11}\)

Câu a :

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

\(\Leftrightarrow x\left(x^2+6x+9\right)-3x=x^3+6x^2+12x+8+1\)

\(\Leftrightarrow x^3+6x^2+9x-3x-x^3-6x^2-12x-8-1=0\)

\(\Leftrightarrow-6x-9=0\) ( dạng ax+b=0)

\(\Rightarrow x=-\dfrac{3}{2}\)

Câu b :

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

\(\Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\)

\(\Leftrightarrow x^2-5x-8-x^2+8x-16=0\)

\(\Leftrightarrow3x-24=0\)

\(\Rightarrow x=8\)

hay wa

x³ + 2x² + x + 2

= (x³ + 2x²) + (x + 2)

= x²(x + 2) + (x + 2)

= (x + 2)(x² + 1)