\(1a,\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)

\(\Leftrightarrow36x=-3\)

\(x=-\frac{1}{12}\)

Vậy ................

\(b,\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

\(\Leftrightarrow\frac{5\left(7x-1\right)}{30}+\frac{30.2x}{30}=\frac{6\left(16-x\right)}{30}\)

\(\Leftrightarrow35x-5+60x=96-6x\)

\(\Leftrightarrow101x=101\)

\(\Leftrightarrow x=1\)

Vậy ....................

Bài 1:

c) \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

\(\Leftrightarrow\frac{8.\left(x-2\right)^2}{8.3}-\frac{3.\left(2x-3\right).\left(2x+3\right)}{3.8}+\frac{4.\left(x-4\right)^2}{4.6}=0\)

\(\Leftrightarrow\frac{8.\left(x^2-4x+4\right)}{24}-\frac{3.\left(4x^2-9\right)}{24}+\frac{4.\left(x^2-8x+16\right)}{24}=0\)

\(\Rightarrow8.\left(x^2-4x+4\right)-3.\left(4x^2-9\right)+4.\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow8x^2-32x+32-\left(12x^2-27\right)+4x^2-32x+64=0\)

\(\Leftrightarrow8x^2-32x+32-12x^2+27+4x^2-32x+64=0\)

\(\Leftrightarrow123-64x=0\)

\(\Leftrightarrow64x=123-0\)

\(\Leftrightarrow64x=123\)

\(\Leftrightarrow x=123:64\)

\(\Leftrightarrow x=\frac{123}{64}.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{123}{64}\right\}.\)

Chúc bạn học tốt!

Bài 1:

a) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2-2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)

\(\Leftrightarrow36x+3=0\)

\(\Leftrightarrow x=12\)

Vậy phương trình có nghiệm là x = 12

Bạn đưa quá nhiều bài 1 lúc nên người ta giải được cũng chẳng ai muốn giải đâu, vì nhìn vào đã thấy ngộp rồi. Kinh nghiệm là muốn được giải quyết nhanh thì chỉ đăng 2-3 bài 1 lúc thôi

Bài 1:

a/ \(11-\left(2x+3\right)=3\left(x-4\right)\)

\(\Leftrightarrow11-2x-3=3x-12\)

\(\Leftrightarrow5x=20\)

\(\Rightarrow x=4\)

b/ \(5\left(2x-3\right)-4\left(5x-7\right)=19-2x\)

\(\Leftrightarrow10x-15-20x+28=19-2x\)

\(\Leftrightarrow8x=-6\)

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

c/

\(\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)

\(\Leftrightarrow2x-3\left(2x+1\right)=x-6x\)

\(\Leftrightarrow x=3\)

d/

\(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)

\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)

\(\Leftrightarrow79x=158\)

\(\Rightarrow x=2\)

e/

\(\frac{2-6x}{5}-\frac{2+3x}{10}=7-\frac{6x+3}{4}\)

\(\Leftrightarrow4\left(2-6x\right)-2\left(2+3x\right)=140-5\left(6x+3\right)\)

\(\Leftrightarrow0=-121\) (vô lý)

Vậy pt vô nghiệm

f/

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

\(\Leftrightarrow3\left(3x+2\right)-\left(3x+1\right)=12x+10\)

\(\Leftrightarrow6x=-5\)

\(\Rightarrow x=-\frac{5}{6}\)

Bài 2:

a/ \(\left(x+1\right)\left(x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=-3\end{matrix}\right.\)

b/

\(\left(x+1\right)^2-4^2=0\)

\(\Leftrightarrow\left(x+1-4\right)\left(x+1+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

c/

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{2}{3}\end{matrix}\right.\)

bn đăng từng câu 1 thôi nhe

Bài 1:

a/ \(x\ne1;2\)

\(\frac{x-2}{\left(x-1\right)\left(x-2\right)}-\frac{7\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}=0\)

\(\Leftrightarrow x-2-7x+7+1=0\)

\(\Leftrightarrow-6x+6=0\)

\(\Rightarrow x=1\) (loại)

Vậy pt vô nghiệm

b/ \(x\ne\frac{3}{2}\)

\(\frac{2x+3}{2x-3}-\frac{3}{2\left(2x-3\right)}-\frac{2}{5}=0\)

\(\Leftrightarrow\frac{10\left(2x+3\right)}{10\left(2x-3\right)}-\frac{15}{10\left(2x-3\right)}-\frac{4\left(2x-3\right)}{10\left(2x-3\right)}=0\)

\(\Leftrightarrow20x+30-15-8x+12=0\)

\(\Leftrightarrow12x+27=0\)

\(\Rightarrow x=-\frac{9}{4}\)

c/ \(x\ne\pm1\)

\(\frac{x+1}{x-1}-\frac{4}{x+1}+\frac{3-x^2}{x^2-1}=0\)

\(\Leftrightarrow\frac{\left(x+1\right)^2}{x^2-1}-\frac{4\left(x-1\right)}{x^2-1}+\frac{3-x^2}{x^2-1}=0\)

\(\Leftrightarrow x^2+2x+1-4x+4+3-x^2=0\)

\(\Leftrightarrow-2x+8=0\)

\(\Rightarrow x=4\)

Bài 1:

d/\(x\ne\pm3\)

\(\frac{x-1}{x+3}-\frac{x}{x-3}+\frac{7x-3}{x^2-9}=0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{x^2-9}-\frac{x\left(x+3\right)}{x^2-9}+\frac{7x-3}{x^2-9}=0\)

\(\Leftrightarrow x^2-4x+3-x^2-3x+7x-3=0\)

\(\Rightarrow0=0\)

Vậy pt có vô số nghiệm \(x\ne\pm3\)

e/ \(x\ne\pm1\)

\(\frac{1}{x+1}+\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}+\frac{3}{x^2-1}=0\)

\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x^2-1\right)\left(x-1\right)}+\frac{3}{x^2-1}=0\)

\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3}{\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=0\)

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

\(\Leftrightarrow x^2+x=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\left(l\right)\end{matrix}\right.\)

Bài 2:

Đặt \(x^2-3=a\) phương trình trở thành:

\(a^2+2a-3=0\)

\(\Leftrightarrow a^2-a+3a-3=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-3=1\\x^2-3=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=0\end{matrix}\right.\)

Bài 3:

a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)

\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)

Vì \(3\ne0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)

b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)

c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)

Chúc bạn học tốt!

Bài 4 xem lại đề nhé bác

Bn làm bài 1 học kì hay 1 tiết

1.

a) 13\(\frac{1}{3}\) : 1\(\frac{1}{3}\) = 26 : (2x - 1)

<=> \(\frac{40}{3}:\frac{4}{3}\) = 13x - 26

<=> 10 + 26 = 13x

<=> 13x = 36

<=> x = \(\frac{36}{13}\)

b) 0,2 : 1\(\frac{1}{5}\) = \(\frac{2}{3}\) : (6x + 7)

<=> \(\frac{1}{5}:\frac{6}{5}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)

<=> \(\frac{1}{6}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)

<=> \(\frac{1}{9}x\) = \(\frac{2}{21}.\frac{1}{6}\) = \(\frac{1}{63}\)

<=> x = \(\frac{1}{7}\)

c) \(\frac{37-x}{x+13}\) = \(\frac{3}{7}\)

<=> (37 - x) . 7 = 3.(x + 13)

<=> 119 - 7x = 3x + 39

<=> -7x - 3x = 39 - 119

<=> -10x = -80

<=> x = 8

d) \(\frac{x-1}{x+5}=\frac{6}{7}\)

<=> 7(x - 1) = 6(x + 5)

<=> 7x - 7 = 6x + 30

<=> 7x - 6x = 30 + 7

<=> x = 37

e)

2\(\frac{2}{\frac{3}{0,002}}\) = \(\frac{1\frac{1}{9}}{x}\)

<=> \(\frac{1501}{750}\) = \(\frac{10}{9}:x\)

<=> x = \(\frac{10}{9}:\frac{1501}{750}\) = \(\frac{2500}{4503}\)

Bài 2. đề sai

Bài 3.

a) 6,88 : x = \(\frac{12}{27}\)

<=> x = 6,88 : \(\frac{12}{27}\)

<=> x = 15,48

b) 8\(\frac{1}{3}\) : \(11\frac{2}{3}\) = 13 : 2x

<=> \(\frac{25}{3}:\frac{35}{3}\) = 13 : 2x

<=> \(\frac{5}{7}=13:2x\)

<=> 2x = \(13:\frac{5}{7}\) = \(\frac{91}{5}\)

<=> x = 9,1

a) \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

\(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{9}{15}-\frac{10}{15}=\frac{-1}{15}\)

\(x=\frac{-1}{15}.\frac{1}{3}\)

\(x=\frac{-1}{45}\)

Vậy x = \(\frac{-1}{45}\)

c) \(\left|2x-1\right|+1=4\)

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

2x-1 = 3 ; -3

TH1: 2.x - 1 = 3

2.x = 3 + 1 = 4

x = 4 : 2 = 2

TH2: 2.x - 1 = -3

2.x = -3 + 1 = -2

x = -2 : 2 = -1

Vậy x \(\in\){ 2 ; -1 }

Ngại làm ấn máy ==