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1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
a) \(\frac{4x-8}{2x^2+1}=0\)
\(\Rightarrow4x-8=0\left(2x^2+1\ne0\right)\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
Vậy x=2
b)
\(\frac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+2\right)}{x-3}=0\)
\(\Rightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
bài 1
\(ĐKXĐ:1+x\ne0\Rightarrow x\ne-1\)
\(\frac{3-7x}{1+x}=\frac{1}{2}\Rightarrow2\left(3-7x\right)=1+x\)
\(\Leftrightarrow6-14x=1+x\\
\Leftrightarrow-14x-x=1-6\\
\Leftrightarrow-15x=-5\\
\Leftrightarrow x=\frac{1}{3}\left(N\right)\)
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 ^-^
Bài 1:
a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)
\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)
\(\Leftrightarrow-5=0\)(vl)
Vậy: \(x\in\varnothing\)
b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\)
hay x=1
Vậy: x=1
c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)
\(\Leftrightarrow2x-72=0\)
\(\Leftrightarrow2\left(x-36\right)=0\)
mà 2>0
nên x-36=0
hay x=36
Vậy: x=36
d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)
\(\Leftrightarrow120x+36=56-64x\)
\(\Leftrightarrow120x+36-56+64x=0\)
\(\Leftrightarrow184x-20=0\)
\(\Leftrightarrow184x=20\)
hay \(x=\frac{5}{46}\)
Vậy: \(x=\frac{5}{46}\)
e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)
\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)
\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)
\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)
\(\Leftrightarrow-23x+29=0\)
\(\Leftrightarrow-23x=-29\)
hay \(x=\frac{29}{23}\)
Vậy: \(x=\frac{29}{23}\)
f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)
\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)
\(\Leftrightarrow2x+8-10x-50-25=0\)
\(\Leftrightarrow-8x-67=0\)
\(\Leftrightarrow-8x=67\)
hay \(x=\frac{-67}{8}\)
Vậy: \(x=\frac{-67}{8}\)
g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)
\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)
\(\Leftrightarrow10-5x-8x-8+12x-30=0\)
\(\Leftrightarrow-x-28=0\)
\(\Leftrightarrow-x=28\)
hay x=-28
Vậy: x=-28
h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)
\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)
\(\Leftrightarrow0x=0\)
Vậy: \(x\in R\)
Bài 2:
a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)
b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)
Từ (1) và (2) suy ra:
\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)
c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)
\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
hay x=-3
Vậy: Tập nghiệm S={-3}
d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)
\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)
\(\Leftrightarrow12-7x=0\)
\(\Leftrightarrow7x=12\)
hay \(x=\frac{12}{7}\)
Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)
e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x
\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow31x=1\)
hay \(x=\frac{1}{31}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)