Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
Bài 1:
a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)
\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)
Suy ra: \(12x-45-12x^2+45x=0\)
\(\Leftrightarrow-12x^2+57x-45=0\)
\(\Leftrightarrow-12x^2+12x+45x-45=0\)
\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)
\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)
mà \(-3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)
b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)
Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)
\(\Leftrightarrow-x^2+16x-39=0\)
\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)
\(\Leftrightarrow x^2-13x-3x+39=0\)
\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)
Vậy: Tập nghiệm S={3;13}
c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)
\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)
Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)
\(\Leftrightarrow-21x^2+26x+11=0\)
\(\Leftrightarrow-21x^2-7x+33x+11=0\)
\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
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!
Vì số lượng bài khá nhiều và mình cũng không có quá nhiều thời gian nên không tránh khỏi sai sót, nếu phát hiện mong bạn thông cảm! Bài của tớ làm khá tắt bước, chỉ nên tham khảo. Bạn có thể tự biểu diễn tập nghiệm được không?
a. \(x+8>3x-1\)
\(\Leftrightarrow-2x>-9\)
\(\Leftrightarrow x< \frac{9}{2}\)
b. \(3x-\left(2x+5\right)\le\left(2x-3\right)\)
\(\Leftrightarrow3x-2x-5\le2x-3\)
\(\Leftrightarrow-x\le2\)
\(\Leftrightarrow x\ge2\)
c. \(\left(x-3\right)\left(x+3\right)< x\left(x+2\right)+3\)
\(\Leftrightarrow x^2-9< x^2+2x+3\)
\(\Leftrightarrow2x>-12\Leftrightarrow x>-6\)
d. \(2\left(3x-1\right)-2x< 2x+1\)
\(\Leftrightarrow6x-2-2x< 2x+1\)
\(\Leftrightarrow2x< 3\)
\(\Leftrightarrow x< \frac{3}{2}\)
e. \(\frac{3-2x}{5}>\frac{2-x}{3}\)
\(\Leftrightarrow3\left(3-2x\right)>5\left(2-x\right)\)
\(\Leftrightarrow9-6x>10-5x\)
\(\Leftrightarrow-x>1\) \(\Leftrightarrow x< -1\)
f. \(\frac{x-2}{6}-\frac{x-1}{3}\le\frac{x}{2}\)
\(\Leftrightarrow x-2-2\left(x-1\right)\le3x\)
\(\Leftrightarrow x-2-2x+2\le3x\)
\(\Leftrightarrow-4x\le0\Leftrightarrow x\ge0\)
g. \(\frac{x+1}{3}>\frac{2x-1}{6}\ge4\)
\(\Leftrightarrow2x+2>2x-1\ge24\)
\(\Leftrightarrow2x+2>2x\ge25\)
\(\Leftrightarrow x\ge\frac{25}{2}\)
h. \(1+\frac{2x+1}{3}>\frac{2x-1}{6}-2\)
\(\Leftrightarrow6+4x+2>2x-1-12\)
\(\Leftrightarrow2x>-25\)
\(\Leftrightarrow x>-\frac{25}{2}\)
i. \(\frac{x+5}{6}-\frac{2x+1}{3}\le\frac{x+3}{2}\)
\(\Leftrightarrow x+5-4x-2\le3x+9\)
\(\Leftrightarrow-6x\le6\)
\(\Leftrightarrow x\ge-1\)
j. \(\frac{5x+4}{6}-\frac{2x-1}{12}\ge4\)
\(\Leftrightarrow10x+8-2x+1\ge48\)
\(\Leftrightarrow8x\ge39\)
\(\Leftrightarrow x\ge\frac{39}{8}\)
Bạn tự biểu diễn nghiệm trên trục số nhé!
a) \(x+8>3x-1\)
\(\Leftrightarrow x-3x>-8-1\)
\(\Leftrightarrow-2x>-9\)
\(\Leftrightarrow x< \frac{9}{2}\)
b) 3x − (2x+5) ≤ (2x−3)
\(\Leftrightarrow3x-2x-5\le2x-3\)
\(\Leftrightarrow3x-2x+2x\le5-3\)
\(\Leftrightarrow3x\le2\)
\(\Leftrightarrow x\le\frac{2}{3}\)
c) (x − 3) (x + 3) < x (x + 2) + 3
\(\Leftrightarrow x^2-9< x^2+2x+3\)
\(\Leftrightarrow x^2-x^2+2x< 9+3\)
\(\Leftrightarrow2x< 12\)
\(\Leftrightarrow x< 6\)
d) 2 (3x − 1) − 2x < 2x + 1
\(\Leftrightarrow6x-2-2x< 2x+1\)
\(\Leftrightarrow6x-2x+2x< 2+1\)
\(\Leftrightarrow6x< 3\)
\(\Leftrightarrow x< \frac{3}{6}\)
e) \(\frac{3-2x}{5}>\frac{2-x}{3}\)
\(\Leftrightarrow\frac{\left(3-2x\right)\times3}{15}>\frac{\left(2-x\right)\times5}{15}\)
\(\Leftrightarrow9-6x>10-5x\)
\(\Leftrightarrow-6x+5x>-9+10\)
\(\Leftrightarrow-x>1\)
\(\Leftrightarrow x< -1\)
f)\(\frac{x-2}{6}-\frac{x-1}{3}\le\frac{x}{2}\)
\(\Leftrightarrow x-2-2\left(x-1\right)\le3x\)
\(\Leftrightarrow x-2-2x+2\le3x\)
\(\Leftrightarrow-4x\le0\)
\(\Leftrightarrow x\ge0\)
g) \(\frac{x+1}{3}>\frac{2x-1}{6}\ge4\)
\(\Leftrightarrow\frac{\left(x+1\right)\cdot2}{6}>\frac{2x-1}{6}\ge\frac{4\cdot6}{6}\)
\(\Leftrightarrow2x+2>2x+1\ge24\)
\(\Leftrightarrow2x+2>2x\ge25\)
\(\Leftrightarrow x\ge\frac{25}{2}\)
h)\(1+\frac{2x+1}{3}>\frac{2x-1}{6}-2\)
\(\Leftrightarrow\frac{1}{6}+\frac{\left(2x+1\right)\cdot2}{6}>\frac{2x-1}{6}-\frac{2\cdot6}{6}\)
\(\Leftrightarrow6+4x+2>2x-1-12\)
\(\Leftrightarrow2x>-21\)
\(\Leftrightarrow x>\frac{-21}{2}\)
i)\(\frac{x+5}{6}-\frac{2x+1}{3}\le\frac{x+3}{2}\)
\(\Leftrightarrow\frac{x+5}{6}-\frac{\left(2x+1\right)\cdot2}{6}\le\frac{\left(x+3\right)\cdot3}{6}\)
\(\Leftrightarrow x+5-4x+2\le3x+9\)
\(\Leftrightarrow-3x-x+4x\le9-5-2\)
\(\Leftrightarrow x\le2\)
j) \(\frac{5x+4}{6}-\frac{2x-1}{12}\ge4\)
\(\Leftrightarrow\frac{\left(5x+4\right)\cdot2}{12}-\frac{2x-1}{12}\ge\frac{4\cdot12}{12}\)
\(\Leftrightarrow10x+8-2x-1\ge48\)
\(\Leftrightarrow10x-2x\ge48-8+1\)
\(\Leftrightarrow8x\ge41\)
\(\Leftrightarrow x\ge\frac{41}{8}\)
Mình không chắc là mình làm đúng đâu. Nhưng có sai sót gì thì cứ nói cho mình biết. Chúc bạn học tốt ^-^
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
a) Qui đồng rồi khử mẫu ta được:
3(3x+2)-(3x+1)=2x.6+5.2
<=> 9x+6-3x-1 = 12x+10
<=> 9x-3x-12x = 10-6+1
<=> -6x = 5
<=> x = -5/6
Vậy ....
b) ĐKXĐ: \(x\ne\pm2\)
Qui đồng rồi khử mẫu ta được:
(x+1)(x+2)+(x-1)(x-2) = 2(x2+2)
<=> x2+3x+2+x2-3x+2 = 2x2+4
<=> x2+x2-2x2+3x-3x = 4-2-2
<=> 0x = 0
<=> x vô số nghiệm
Vậy x vô số nghiệm với x khác 2 và x khác -2
c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)
Vậy ......
d) (x+1)2-4(x2-2x+1) = 0
<=> x2+2x+1-4x2+8x-4 = 0
<=> -3x2+10x-3 = 0
giải phương trình
