a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)

Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)

\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)

\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)

\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)

\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)

\(< =>3072-107x=\frac{38x-684}{5}\)

\(< =>\left(3072-107x\right)5=38x-684\)

\(< =>15360-535x-38x-684=0\)

\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)

nghệm xấu thế 

\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)

\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)

\(< =>993-33x-11x-415=0\)

\(< =>578=44x< =>x=\frac{289}{22}\)

Bài 1: 

b) Phương trình đã cho tương đương với phương trình:

\(\frac{8\left(x+22\right)-55\left(7x+149\right)-6\left(x+12\right)}{45}=\frac{9\left(x+35\right)+2\left(x+50\right)}{45}\)

\(\Leftrightarrow44x=-1056\)

\(\Leftrightarrow x=-24\)

Vậy x=-24 là nghiệm của phương trình

c) Phương trình đã cho tương đương với phương trình:

\(\frac{3x+6}{70}-\frac{x+4}{24}=\frac{32x+19}{60}+\frac{2}{3}\)

\(\Leftrightarrow12\left(3x+6\right)-35\left(x+4\right)=14\left(32x+19\right)+560\)

\(\Leftrightarrow-447x=894\)

\(\Leftrightarrow x=-2\)

Vậy x=-2 là nghiệm của phương trình

a) \(x^3-2x^2-5x+6=0\)

\(x^3-x^2-x^2+x-6x+6=0\)

\(x^2\left(x-1\right)-x\left(x-1\right)-6\left(x-1\right)=0\)

\(\left(x-1\right)\left(x^2-x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x^2-x-6=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x^2-2x+3x-6=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\\left(x+3\right)\left(x-2\right)=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\left\{2;-3\right\}\end{cases}}\)

\(a,x^3-2x^2-5x+6=0\)

\(\Leftrightarrow\left(x^3-x^2\right)-\left(x^2-x\right)-\left(6x-6\right)=0\)

\(\Leftrightarrow x^2\left(x-1\right)-x\left(x-1\right)-6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^2-3x\right)+\left(2x-6\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow x-1=0\left(h\right)x+2=0\left(h\right)x-3=0\)

\(\Leftrightarrow x=1\left(h\right)x=-2\left(h\right)x=3\)

Vậy \(x\in\left\{-2;1;3\right\}\)

P/S: (h) là hoặc nhé

\(b,\left|5-3x\right|=3x-5\)

*Nếu \(x\ge\frac{5}{3}\)thì

\(3x-5=3x-5\)Luôn đúng \(\forall x\ge\frac{5}{3}\)

*Nếu \(x< \frac{5}{3}\)thì

\(5-3x=3x-5\)

\(\Leftrightarrow6x=10\)

\(\Leftrightarrow x=\frac{5}{3}\)(loại vì ko thỏa mãn khoảng đag xét)

Vậy \(x\ge\frac{5}{3}\)

Cách khác : dùng tính chất của trị tuyệt đối

\(\left|5-3x\right|=3x-5\)

Vì \(VT\ge0\Rightarrow VP\ge0\)

                  \(\Leftrightarrow3x-5\ge0\)

                   \(\Leftrightarrow x\ge\frac{5}{3}\)

Vậy ...........

A/  \(2\left(5x-3\right)=7x-18.\)

\(10x-6=7x-18\)

\(10-7x=6-18\)

\(3x=-12\)

\(x=-\frac{12}{3}=4\)

\(\Rightarrow S=\left\{4\right\}\)

B/  \(3x\left(x-2\right)+2x-4=0\)

\(3x\left(x-2\right)+2\left(x-2\right)=0\)

\(\left(x-2\right)\left(3x+2\right)=0\)

\(\orbr{\begin{cases}x-2=0\Rightarrow x=2\\3x+2=0\Rightarrow3x=-2\Rightarrow x=-\frac{2}{3}\end{cases}}\)

\(\Rightarrow S=\left\{2;-\frac{2}{3}\right\}\)

C/  \(\frac{x+2}{3}\frac{x-3}{2}=\frac{x+5}{4}\)

\(\frac{\left(x+2\right)\left(x-3\right)}{3.2}=\frac{x+5}{4}\)

\(\frac{x^2-3x+2x-6}{6}=\frac{x+5}{4}\)

\(\frac{x^2-x-6}{6}=\frac{x+5}{4}\)

\(\frac{2\left(x^2-x-6\right)}{12}=\frac{3\left(x+5\right)}{12}\)

\(\frac{2x^2-2x-12}{12}=\frac{3x+15}{12}\)

\(\Rightarrow2x^2-2x-12=3x+15\)

(chuyển vế r làm tiếp)

Bài 1 : 

\(a,2\left(5x-3\right)=7x-18\)

\(\Leftrightarrow10x-6=7x-18\)

\(\Leftrightarrow10x-7x=6-18\)

\(\Leftrightarrow3x=-12\)

\(\Leftrightarrow x=-4\)

PT có nghiệm S = { -4 }

\(b,3x\left(x-2\right)+2x-4=0\)

\(\Leftrightarrow3x^2-6x+2x-4=0\)

\(\Leftrightarrow3x^2-4x-4=0\)

\(\Leftrightarrow3x^2-6x+2x-4=0\)

\(\Leftrightarrow3x\left(x-2\right)+2\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x+2=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=2\end{cases}}\)

KL : ............

\(c,\frac{x+2}{3}-\frac{x-3}{2}=\frac{x+5}{4}\)

\(\Leftrightarrow\frac{4\left(x+2\right)}{12}-\frac{6\left(x-3\right)}{12}=\frac{3\left(x+5\right)}{12}\)

\(\Leftrightarrow4x+8-6x+18=3x+15\)

\(\Leftrightarrow4x-6x-3x=-8-18+15\)

\(\Leftrightarrow x=-9\)

KL : .......

c/  \(\frac{5}{x-3}+\frac{2}{x-2}=\frac{5}{\left(x-3\right)\left(x-2\right)}.\)

\(\frac{5\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}+\frac{2\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}=\frac{5}{\left(x-3\right)\left(x-2\right)}\)

\(\frac{5x-10}{\left(x-3\right)\left(x-2\right)}+\frac{2x-6}{\left(x-3\right)\left(x-2\right)}=\frac{5}{\left(x-3\right)\left(x-2\right)}\)

\(\frac{5x-10+2x-6}{\left(x-3\right)\left(x-2\right)}=\frac{5}{\left(x-3\right)\left(x-2\right)}\)

\(\frac{7x-16}{\left(x-3\right)\left(x-2\right)}=\frac{5}{\left(x-3\right)\left(x-2\right)}\)

\(\Rightarrow7x-16=5\)

\(\Rightarrow7x=5+16\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=\frac{21}{7}=3\)

\(\Rightarrow S=\left\{3\right\}\)

D/ \(\frac{7x-2}{3}-2x< 5-\frac{x-2}{4}\)

\(\frac{7x-2}{3}-\frac{2x.3}{3}< \frac{5.4}{4}-\frac{x-2}{4}\)

\(\frac{7x-2}{3}-\frac{6x}{3}< \frac{20}{4}-\frac{x-2}{4}\)

\(\frac{7x-2-6x}{3}< \frac{20-x+2}{4}\)

\(\frac{x-2}{3}< \frac{22-x}{4}\)

\(\frac{4\left(x-2\right)}{12}< \frac{3\left(22-x\right)}{12}\)

\(\frac{4x-8}{12}< \frac{66-3x}{12}\)

\(\Rightarrow4x-8=66-3x\)

\(\Rightarrow4x+3x=66+8\)

\(\Rightarrow7x=74\)

\(\Rightarrow x=\frac{74}{7}\)

\(\Rightarrow S=\left\{\frac{74}{7}\right\}\)

a , 2x -3 = 5x + 6

    2x -5x=6+3

    -3x = 9

     x =9 :(-3)

   x= -3

a) 2x-5x=3+6

-3x=9

x=-3

vậy........

b)(2x+1).(3x-2)-(5x-8).(2x+1)=0

(2x+1).(3x-2-2x-1)=0

(2x-1).(x-3)=0

==>x=1/2 ; x=3

c)(2x+1).5-(7x+5)=(2x-2).3

10x+5-7x-5=6x-6

3x=6x-6

3x-6x=6

-3x=6

x=-2

a) 2x - 3 = 5x + 6

<=> -3x = 9

<=> x = -3

b) (2x + 1).(3x - 2) = (5x - 8).(2x + 1)

<=> 6x² - 4x + 3x - 2 = 10x² + 5x - 16x -8

<=> -4x² - 10x + 6 = 0

<=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)

c) \(\frac{2\text{x}+1}{3}-\frac{7\text{x}+5}{15}=\frac{2\text{x}-2}{5}\)

<=> \(\frac{5.\left(2\text{x}+1\right)}{5.3}-\frac{7\text{x}+5}{15}=\frac{3.\left(2\text{x}-2\right)}{3.5}\)

<=> 10x + 5 - 7x + 5 - 6x + 6 = 0

<=> -3x + 16 = 0

<=> -3x = -16

<=> x = \(\frac{16}{3}\)

d) \(\frac{3x}{x-2}-\frac{x}{x-5}=\frac{3\text{x}}{\left(x-2\right).\left(5-x\right)}\)

<=> \(\frac{3\text{x}\left(x-5\right)-x\left(x-2\right)}{\left(x-2\right).\left(5-x\right)}=\frac{3\text{x}}{\left(x-2\right).\left(5-x\right)}\)

<=> 3x² - 15x - x² + 2x - 3x = 0

*Câu e dễ quá, bạn tự làm nhé :v*

a/ Đơn giản, phân tích mẫu số thứ 3 thành nhân tử rồi quy đồng, ko có gì khó cả, chắc bạn tự làm được

b/ Đặt \(\left(x+1\right)^2=t\ge0\)

\(\frac{t+6}{t+2}=t+3\Leftrightarrow t+6=\left(t+2\right)\left(t+3\right)\)

\(\Leftrightarrow t^2+4t=0\Rightarrow\orbr{\begin{cases}t=0\\t=-4\left(l\right)\end{cases}}\) \(\Rightarrow x=-1\)

c/ ĐKXĐ: bla bla bla...

Nhận thây \(x=0\) không phải nghiệm, phương trình tương đương:

\(\frac{2}{3x+\frac{2}{x}-1}-\frac{7}{3x+\frac{2}{x}+5}=1\)

Đặt \(3x+\frac{2}{x}-1=t\)

\(\frac{2}{t}-\frac{7}{t+6}=1\)

\(\Leftrightarrow2\left(t+6\right)-7t=t\left(t+6\right)\)

\(\Leftrightarrow t^2+11t-12=0\Rightarrow\orbr{\begin{cases}t=1\\t=-12\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x+\frac{2}{x}-1=1\\3x+\frac{2}{x}-1=-12\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}3x^2-2x+2=0\\3x^2+11x+2=0\end{cases}}\)

Bấm máy