a.. \(\frac{7-8x}{6}=\frac{-4+2x}{5}\)
\(b.\frac{1-3:x}{8}=\frac{8}{1-3:x}\)
c. \(\left(x+1\right).\left(x-2\right)\ge0\)
h. \(\left(x+1\right).\left(x-3\right)\le0\)
AI làm nahnh mih tích cho nha đang rất rất cần
a.. \(\frac{7-8x}{6}=\frac{-4+2x}{5}\)
\(b.\frac{1-3:x}{8}=\frac{8}{1-3:x}\)
c. \(\left(x+1\right).\left(x-2\right)\ge0\)
h. \(\left(x+1\right).\left(x-3\right)\le0\)
AI làm nhanh mih tích mih tích cho nha đang rất rất cần
a) \(\frac{7-8x}{6}=\frac{-4+2x}{5}\)
=> \(\left(7-8x\right).5=6\left(-4+2x\right)\)
=> 35 - 40x = -24 + 12x
=> 35 + 24 = 12x + 40x
=> 52x = 59
=> x = 59/52
b) \(\frac{1-3:x}{8}=\frac{8}{1-3:x}\)
=> (1 - 3: x)2 = 82
=> \(\orbr{\begin{cases}1-3:x=8\\1-3:x=-8\end{cases}}\)
=> \(\orbr{\begin{cases}3:x=-7\\3:x=9\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{3}{7}\\x=\frac{1}{3}\end{cases}}\)
c) \(\left(x+1\right)\left(x-2\right)\ge0\)
=> \(\hept{\begin{cases}x+1\ge0\\x-2\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-2\le0\end{cases}}\)
=> \(\hept{\begin{cases}x\ge-1\\x\ge2\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\le2\end{cases}}\)
=> \(-1\le x\le2\)
h) \(\left(x+1\right)\left(x-3\right)\le0\)
=> \(\hept{\begin{cases}x+1\ge0\\x-3\le0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-3\ge0\end{cases}}\)
=> \(\hept{\begin{cases}x\ge-1\\x\le3\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\ge3\end{cases}}\) (loại)
= \(-1\le x\le3\)
Mình làm tiếp bạn Edogawa Conan ! Hai bài a và c bạn ấy làm đúng rồi đó !
b, \(\frac{1-3\text{ : }x}{8}=\frac{8}{1-3\text{ : }x}\)
\(\Rightarrow\text{ }\left(1-3\text{ : }x\right)\text{ x }\left(1-3\text{ : }x\right)=8\text{ x }8\)
\(\text{ }\left(1-3\text{ : }x\right)^2=8^2\)
\(\Rightarrow\text{ }1-3\text{ : }x=8\)
\(3\text{ : }x=1-8\)
\(3\text{ : }x=-7\)
\(x=-\frac{7}{3}\)
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 bất phương trình sau
1 , \(\left(2x+3\right)\left(5x-7\right)\ge0\)
2 , \(\left(3-2x\right)\left(4x+3\right)< 0\)
3 , \(\left(2x+5\right)\left(3-x\right)\left(5x-1\right)\le0\)
4 , \(x^2-3x+2< 0\)
5 , \(-x^2+12x+13>0\)
6 , \(x^2+6x+9\le0\)
7 , \(\frac{x+2}{3x+1}>\frac{x-2}{2x-1}\)
8 , \(\frac{1}{x+2}< \frac{3}{x-3}\)
9 , \(\frac{5x-6}{2x-5}\le6\)
10 , \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)
Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
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 :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
1) Giải các pt sau:
a) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
b) \(\frac{3x-2}{6}-5=\frac{3-2\left(x+7\right)}{4}\)
c) \(\frac{x+8}{6}-\frac{2x-5}{5}=\frac{x-1}{3}-x+7\)
d) \(\frac{7x}{8}-5\left(x-9\right)=\frac{2x+1,5}{6}\)
e) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
f) \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)
⇔ 3(x - 3) = 90 - 5(1 - 2x)
⇔ 3x - 9 = 90 - 5 + 10x
⇔ 3x - 10x = 90 - 5 + 9
⇔ -7x = 94
⇔ x = \(\frac{-94}{7}\)
S = { \(\frac{-94}{7}\) }
b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)
⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)
⇔ 6x - 4 - 60 = 9 - 6x - 42
⇔ 6x + 6x = 9 - 42 + 60 + 4
⇔ 12x = 31
⇔ x = \(\frac{31}{12}\)
S = { \(\frac{31}{12}\) }
c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7
⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210
⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210
⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40
⇔ 13x = 150
⇔ x = \(\frac{150}{13}\)
S = { \(\frac{150}{13}\) }
d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)
⇔ 21x - 120(x - 9) = 4(2x + 1,5)
⇔ 21x - 120x + 1080 = 8x + 6
⇔ 21x - 120x - 8x = 6 - 1080
⇔ -107x = -1074
⇔ x = \(\frac{1074}{107}\)
S = { \(\frac{1074}{107}\) }
e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5
⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840
⇔ 140x -140+56 -294x+42= 96x+48 -840
⇔ 140x -294x -96x = 48 -840 -42 -56+140
⇔ -250x = -750
⇔ x = 3
S = { 3 }
f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)
⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x
⇔ 4x+4+18x+9 = 4x+6x+6+7+12x
⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4
⇔ 0x = 0
S = R
Chúc bạn học tốt !
giải này:
A=\(\frac{2x}{x+1}-\frac{x}{x-3}=\frac{8}{x^2-2x-3}\) MTC (x+1)(x-3)
A=\(\frac{2x\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}-\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-3\right)}=\frac{8}{\left(x+1\right)\left(x-3\right)}\)
=>A=2x(x-3)-x(x+1)=8
A=2x2-6x-x2-x=8
A=x2-7x=8
A=x2-7x-8x=0
a-b+c=1+7-8=0
x1=-1
x2=\(-\frac{c}{a}\)=8
Giải các phương trình,bất phương trình:
c,\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
d,\(\frac{4}{-25x^2+20x-3}=\frac{3}{5x-1}-\frac{2}{5x-3}\)
e,\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{2}{x^2-4x+3}=0\)
g,\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)
h,\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
i,\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
k,\(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)
l,\(\left(x^2-2x+1\right)-4=0\)
m,\(4x^2+4x++1=x^2\)
Xin đáy ai giúp mình đi
Bài 1:tính giá trị biểu thức sau:
\(B=\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).\left(1-\frac{1}{36}\right)...\left(1-\frac{1}{1326}\right)\)
Bài 2 tìm x biết
\(a,1\frac{1}{30}:\left(24\frac{1}{6}-24\frac{1}{5}\right)-\frac{1\frac{1}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(-1\frac{1}{15}\right):\left(8\frac{1}{5}-8\frac{1}{3}\right)\)
\(b,\left|3x-\frac{2}{5}\right|=\frac{1}{35}+\left|\frac{-3}{7}\right|\)
\(c,-3\frac{1}{2}-\left(2x+\frac{1}{2}\right):\left(-\frac{3}{7}\right)=0\)
\(d,\left(x-4\right).\left(x+2\right)\le0\)với x thuộc z
giúp mk với