Giải PT
1.
a. \(2.\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\)
c. \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
giải các hệ BPT sau:
a) \(\left\{{}\begin{matrix}5x-2>4x+5\\5x-4< x+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x+1>3x+4\\5x+3\ge8x-9\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\frac{5x+2}{3}\ge4-x\\\frac{6-5x}{13}< 3x+1\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}6x+\frac{5}{7}< 4x+7\\\frac{8x+3}{2}< 2x+5\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}15x-2>2x+\frac{1}{3}\\2\left(x-4\right)< \frac{3x-14}{2}\end{matrix}\right.\)
g) \(\left\{{}\begin{matrix}x-1\le2x-3\\3x< x+5\\5-3x\le2x-6\end{matrix}\right.\)
h) \(\left\{{}\begin{matrix}2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\)
j) \(\left\{{}\begin{matrix}\frac{3x+1}{2}-\frac{3-x}{3}\le\frac{x+1}{4}-\frac{2x-1}{3}\\3-\frac{2x+1}{5}>x+\frac{4}{3}\end{matrix}\right.\)
Giải các phương trình sau:
a)\(\frac{\left(9x-0.7\right)}{4}-\frac{\left(5x-1.5\right)}{7}=\frac{\left(7x-1.1\right)}{3}-\frac{5\left(0.4-2x\right)}{6}\)
b)\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}\)
c)\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=-\frac{7}{6\left(x+5\right)}\)
d)\(\frac{8x^2}{3\left(1-4x\right)^2}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
Bài 1:Giải phương trình
a)\(10x^2-5x\left(2x+3\right)=15\)
b)\(3x-7-\left(3-4x\right)\left(2x+1\right)=4x\left(2x-7\right)\)
c)\(\left(4x-5\right)^2-\left(7-2x\right)=4\left(2x-4\right)^2+6x\)
Bài 2:Giải phương trình
a)\(\frac{3\left(x-1\right)}{2}+4=\frac{2x}{3}+\frac{4-5x}{6}\)
b)\(\frac{4-x}{7}-\frac{1}{7}\left(\frac{7+3x}{9}+\frac{5-2x}{2}\right)=4-\frac{4x}{3}\)
c)\(\frac{2}{9}\left(2x-5\right)-\frac{5}{3}\left[\left(x-2\right)-\frac{7}{12}\right]=\frac{3}{4}\left(x-3\right)\)
Bài 3:Giải phương trình
a)\(\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\)
b)\(2x\left(x-3\right)+5\left(x-3\right)=0\)
c)\(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
Bài 4:Tìm m để phương trình sau có nghiệm bằng 7:\(\left(2m-5\right)x-2m^2+8=43\)
Bài 5:Giải phương trình
a)\(\left(2x-1\right)^2-\left(2x+1\right)^2=0\)
b)\(\frac{1}{27}\left(x-3\right)^3-\frac{1}{125}\left(x-5\right)^3=0\)
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
Bài 3 : Giải các phương trình sau bằng cách đưa về dạng ax+b=0 :
a) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
b) \(4\left(0,5-1,5x\right)=-\frac{5x-6}{3}\)
c) \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
d) \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
e) \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right)-\frac{1}{3}\left(x+2\right)\)
a) Ta có: \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-5x+45-\frac{20x+1,5}{6}=0\)
\(\Leftrightarrow\frac{21x}{24}-\frac{120x}{24}+\frac{1080}{24}-\frac{4\left(20x+1,5\right)}{24}=0\)
\(\Leftrightarrow-99x+1080-4\left(20x+1,5\right)=0\)
\(\Leftrightarrow-99x+1080-80x-6=0\)
\(\Leftrightarrow1074-179x=0\)
\(\Leftrightarrow179x=1074\)
hay x=6
Vậy: x=6
b) Ta có: \(4\left(0,5-1,5x\right)=-\frac{5x-6}{3}\)
\(\Leftrightarrow2-6x=\frac{6-5x}{3}\)
\(\Leftrightarrow\frac{3\left(2-6x\right)}{3}-\frac{6-5x}{3}=0\)
\(\Leftrightarrow6-18x-6+5x=0\)
\(\Leftrightarrow-13x=0\)
mà -13≠0
nên x=0
Vậy: x=0
c) Ta có: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
\(\Leftrightarrow\frac{6\left(x+4\right)}{30}+\frac{30\left(-x+4\right)}{30}-\frac{10x}{30}+\frac{15\left(x-2\right)}{30}=0\)
\(\Leftrightarrow6\left(x+4\right)+30\left(4-x\right)-10x+15\left(x-2\right)=0\)
\(\Leftrightarrow6x+24+120-30x-10x+15x-30=0\)
\(\Leftrightarrow-19x+114=0\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: x=6
d) Ta có: \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
\(\Leftrightarrow\frac{21\left(4x+3\right)}{105}-\frac{15\left(6x-2\right)}{105}-\frac{35\left(5x+4\right)}{105}-\frac{315}{105}=0\)
\(\Leftrightarrow84x+63-90x+30-175x-140-315=0\)
\(\Leftrightarrow-181x-362=0\)
\(\Leftrightarrow-181x=362\)
hay x=-2
Vậy: x=-2
e) Ta có: \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right)-\frac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\frac{x+3}{4}=3-\frac{x+1}{2}-\frac{x+2}{3}\)
\(\Leftrightarrow\frac{3\left(x+3\right)}{12}-\frac{36}{12}+\frac{6\left(x+1\right)}{12}+\frac{4\left(x+2\right)}{12}=0\)
\(\Leftrightarrow3x+9-36+6x+6+4x+8=0\)
\(\Leftrightarrow13x-13=0\)
\(\Leftrightarrow13x=13\)
hay x=1
Vậy: x=1
Bài 1. Giải các phương trình sau
1) \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}-2x\)
2) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
3) \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\)
4) \(\frac{2x+3}{3}=\frac{5-4x}{2}\)
5) \(\frac{5x+3}{12}=\frac{1+2x}{9}\)
6) \(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
7) \(\frac{3\left(x-3\right)}{4}+\frac{4x-10,5}{10}=\frac{3\left(x+1\right)}{5}+6\)
8) \(\frac{2\left(3x+1\right)+1}{4}-5=\frac{2 \left(3x-1\right)}{5}-\frac{3x+2}{10}\)
9) \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
10) \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
Giải các phương trình sau:
a) \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
b) \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
a) \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
=> \(\frac{x+2+98}{98}+\frac{x+3+97}{97}=\frac{x+4+96}{96}+\frac{x+5+95}{95}\)
=> \(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
=> \(\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Ta có : \(\frac{1}{98}+\frac{1}{97}\ne\frac{1}{96}+\frac{1}{95}\) => \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
=> \(x+100=0\)
=> \(x=-100\)
Giải các phương trình sau:
a) \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
b) \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
Nhiều thế bạn tách từng câu ra mik giải cho (olm ko dc trừ điểm câu này của e)
Phần b bạn tự làm nhé, chỉ cần quy đồng lên lấy MC = 105 là được mà
Phần a mình giải ntn:
PT \(\Leftrightarrow\) \(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\)
\(\Leftrightarrow x=-100\)
Giải bất phương trình:
a) \(\left(x-3\right)^2< x^2-5x+4\)
b) \(\left(x-3\right)\left(x+3\right)\le\left(x+2\right)^2+3\)
c)\(\frac{4x-5}{7}>\frac{7-x}{5}\)
d) \(\frac{2x+1}{2}+3\ge\frac{3-5x}{3}-\frac{4x+1}{4}\)
Tìm điều kiện xác định rồi giải các phương trình sau:
a) \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
b) \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
c) \(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
Help me!
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi