Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)
<=> \(-\frac{4}{3}x=-\frac{59}{24}\)
<=> \(x=\frac{59}{32}\)
Vậy S = { 59/32}
b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)
<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)
<=> \(-x=-8\)
<=> x = 8
Vậy S = { 8 }
mình làm câu cuối thôi nhé , những câu còn lại bạn tự làm đi , dễ mà :)))) chỉ cần quy đồng mẫu lên là được
\(=\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}\)
\(=\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)
\(=\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)
\(=\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)
Vì \(\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)\) luôn khác 0
<=> x + 59 = 0
<=> x=-59
1) Ta có: x-4=2x+4
\(\Leftrightarrow x-4-2x-4=0\)
\(\Leftrightarrow-x-8=0\)
\(\Leftrightarrow-x=8\)
hay x=-8
Vậy: S={8}
2) Ta có: \(\frac{2x-1}{2}-\frac{x}{3}=x-\frac{x}{6}\)
\(\Leftrightarrow\frac{3\left(2x-1\right)}{6}-\frac{2x}{6}=\frac{6x}{6}-\frac{x}{6}\)
\(\Leftrightarrow3\left(2x-1\right)-2x-6x+x=0\)
\(\Leftrightarrow6x-3-2x-6x+x=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
hay x=-3
Vậy: S={-3}
3) ĐKXĐ: \(x\notin\left\{\frac{-1}{2};3\right\}\)
Ta có: \(\frac{x+3}{2x+1}-\frac{x}{x-3}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x-3\right)}{\left(2x+1\right)\left(x-3\right)}-\frac{x\left(2x+1\right)}{\left(x-3\right)\left(2x+1\right)}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)
Suy ra: \(x^2-9-\left(2x^2+x\right)-3x^2-x-9=0\)
\(\Leftrightarrow-2x^2-x-18-2x^2-x=0\)
\(\Leftrightarrow-4x^2-2x-18=0\)
\(\Leftrightarrow-4\left(x^2+\frac{1}{2}x+\frac{4}{5}\right)=0\)
\(\Leftrightarrow x^2+\frac{1}{2}x+\frac{4}{5}=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{1}{4}+\frac{1}{16}+\frac{59}{80}=0\)
\(\Leftrightarrow\left(x+\frac{1}{4}\right)^2+\frac{59}{80}=0\)(vô lý)
Vậy: S=\(\varnothing\)
4) Ta có: \(\frac{2x}{3}+\frac{2x-1}{6}=4-\frac{x}{3}\)
\(\Leftrightarrow\frac{4x}{6}+\frac{2x-1}{6}=\frac{24}{6}-\frac{2x}{6}\)
\(\Leftrightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow6x-1-24+2x=0\)
\(\Leftrightarrow8x-25=0\)
\(\Leftrightarrow8x=25\)
hay \(x=\frac{25}{8}\)
Vậy: \(S=\left\{\frac{25}{8}\right\}\)
a) \(\frac{1}{x+2}+\frac{2}{x+3}=\frac{6}{x+4}\)
ĐKXĐ \(x\ne-2,-3,-4\)
=> \(\frac{1}{x+2}+\frac{2}{x+3}-\frac{6}{x+4}=0\)
=> \(\frac{3x+7}{\left(x+2\right)\left(x+3\right)}-\frac{6}{x+4}=0\)
=> \(\frac{\left(3x+7\right)\left(x+4\right)-6\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+4\right)}=0\)
=> (3x + 7)(x + 4) - 6(x2 + 5x + 6) = 0
=> 3x2 + 19x + 28 - 6x2 - 30x - 36 = 0
=> -3x2 - 11x - 8 = 0
=> -3x2 - 3x - 8x - 8 = 0
=> -3x(x + 1) - 8(x + 1) = 0
=> (x + 1)(-3x - 8) = 0
=> \(\orbr{\begin{cases}x=-1\\x=-\frac{8}{3}\end{cases}}\)
Vậy ...
b) Thiếu dữ liệu cuả đề
c) \(\frac{6x+22}{x+2}-\frac{2x+7}{x+3}=\frac{x+4}{x^2+5x+6}\)
ĐKXĐ \(x\ne-2;-3\)
=> \(\frac{\left(6x+22\right)\left(x+3\right)-\left(x+2\right)\left(2x+7\right)}{\left(x+2\right)\left(x+3\right)}=\frac{x+4}{\left(x+2\right)\left(x+3\right)}\)
=> \(6x^2+40x+66-x\left(2x+7\right)-2\left(2x+7\right)=x+4\)
=> \(6x^2+40x+66-2x^2-7x-4x-14=x+4\)
=> 4x2 + 29x + 52 = x + 4
=> 4x2 + 29x + 52 - x - 4 = 0
=> 4x2 + 28x + 48 = 0
=> 4(x2 + 7x + 12) = 0
=> x2 + 7x +12 = 0
=> x2 + 3x + 4x + 12 = 0
=> x(x + 3) + 4(x + 3) = 0
=> (x + 3)(x + 4) = 0
=> \(\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)
Mà \(x\ne-2,-3\)nên x = -3 loại
Vậy x = -4
câu a tự quy đồng cùng mẫu rồi làm thôi :"))
b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)
\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)
Đặt \(x^2-x=k\), ta có:
\(k.\left(k-2\right)=24\)
\(\Leftrightarrow k^2-2k+1=25\)
\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)
\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)
c)\(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)
\(\Leftrightarrow x^3.\left(x+2\right)+2x.\left(x+2\right)+3.\left(x^2-2^2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3-x^2+x^2-x+6x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right).\left(x-1\right).\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)
p/s: bn tự kết luận nha :))
a) \(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow-21x=3x-60\)
\(\Leftrightarrow24x=60\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{5}{2}\right\}\)
b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{\left(8x-3\right)-2\left(3x-2\right)}{4}=\frac{2\left(2x-1\right)+\left(x+3\right)}{4}\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)
c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
\(\Leftrightarrow\frac{15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)}{30}=0\)
\(\Leftrightarrow15x-15-2x-2-10x+65=0\)
\(\Leftrightarrow3x+48=0\)
\(\Leftrightarrow x=-16\)
Vậy tập nghiệm của phương trình là \(S=\left\{-16\right\}\)
d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
\(\Leftrightarrow\frac{9\left(3-x\right)+16\left(5-x\right)}{24}=\frac{12\left(1-x\right)-48}{24}\)
\(\Leftrightarrow27-9x+80-16x=12-12x-48\)
\(\Leftrightarrow-25x+107=-12x-36\)
\(\Leftrightarrow-13x+143=0\)
\(\Leftrightarrow x=11\)
Vậy tập nghiệm của phương trình là \(S=\left\{11\right\}\)
Đkxđ: \(\left\{{}\begin{matrix}x\ne0\\x\ne2\\x\ne-2\end{matrix}\right.\)
\(\frac{12}{x^2-4}+\frac{1}{2x-x^2}=\frac{4+x}{x\left(x+2\right)}\)\(\Leftrightarrow \dfrac{12x}{x(x-2)(x+2)}+\dfrac{(x+2)}{x(2-x)(x+2}-\dfrac{4x(x-2)}{x(x+2)(x-2)}=0\)
\(\Leftrightarrow \dfrac{12x-(x+2)-4x(x-2)}{x(x-2)(x+2)}=0\)
\(\Leftrightarrow -4x^2+11x=0\)\(\Leftrightarrow x\left(-4x+11\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\-4x+11=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{11}{4}\end{matrix}\right.\)
KL:........................