x + \(3\frac{1}{2}+x=24\frac{1}{4}\)
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\(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
\(\Leftrightarrow\frac{\frac{5x+6-2x}{5}}{14}-\frac{x+4}{24}=\frac{\frac{35x+10+9-3x}{5}}{12}+\frac{2}{3}\)
\(\Leftrightarrow\frac{\frac{3x+6}{5}}{14}-\frac{x+4}{24}=\frac{\frac{32x+19}{5}}{12}+\frac{2}{3}\)
\(\Leftrightarrow\left(\frac{3x+6}{5}\cdot\frac{1}{14}\right)-\frac{x+4}{24}=\left(\frac{32x+19}{5}\cdot\frac{1}{12}\right)+\frac{2}{3}\)(CHIA CHO 14 LÀ NHÂN NGHỊCH ĐẢO VỚI 1/14,) (CHIA CHO 12 LÀ NHÂN NGHỊCH ĐẢO VỚI 1/12)\(\Leftrightarrow\frac{3x+6}{70}-\frac{x+4}{24}-\frac{32x+19}{60}-\frac{2}{3}=0\)\(\Leftrightarrow\frac{12\left(3x+6\right)-35\left(x+4\right)-14\left(32x+19\right)-2\cdot280}{840}=0\)
\(\Leftrightarrow12\left(3x+6\right)-35\left(x+4\right)-14\left(32x+19\right)-560=0\)
\(\Leftrightarrow36x+72-35x-140-448x-266-560=0\)
\(\Leftrightarrow-447x-894=0\Leftrightarrow x=\frac{-894}{447}=-2\)(NHẬN)
Vậy tập nghiệm của phương trình là : S = { -2 }
tk cho mk nka ! ! ! th@nks ! ! !
\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)
\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)
\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)
\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)
=> x - 29 = 0
=> x = 29.
a) \(x+4\frac{1}{5}=5\frac{2}{3}\times10\frac{1}{2}\)
\(x+\frac{21}{5}=\frac{17}{3}\times\frac{21}{2}\)
\(x+\frac{21}{5}=\frac{119}{2}\)
\(x=\frac{119}{2}-\frac{21}{5}\)
\(x=\frac{553}{10}\)
b) \(\frac{24}{5}-x=2\frac{1}{4}:1\frac{3}{4}\)
\(\frac{24}{5}-x=\frac{9}{4}:\frac{7}{4}\)
\(\frac{24}{5}-x=\frac{9}{4}.\frac{4}{7}\)
\(\frac{24}{5}-x=\frac{9}{7}\)
\(x=\frac{24}{5}-\frac{9}{7}\)
\(x=\frac{123}{35}\)
a, (x-15):5+22=24
( x - 15 ) : 5 = 2
x-15 = 10
x = 25
c,chia cả tử và mẫu cho x,sau đó đặt 3x+2/x=t
các câu còn lại hiện chưa giải đc vì chưa có giấy nháp,lúc nào rảnh mình chỉ cho cách làm
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
<=> \(\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]-24=0\)
<=> \(\left(x^2+x\right)\left(x^2+2x-x-2\right)-24=0\)
<=> \(\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
Đặt t = x2 + x
<=> t(t - 2) - 24 = 0
<=> t2 - 2t - 24 = 0
<=> t2 - 6t + 4t - 24 = 0
<=> (t + 4)(t - 6) = 0
<=> \(\orbr{\begin{cases}x^2+x+4=0\\x^2+x-6=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x^2+x+\frac{1}{4}\right)+\frac{15}{4}=0\\x^2+3x-2x-6=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\\\left(x-2\right)\left(x+3\right)=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Vậy S = {2; -3}
(lưu ý: thay "ktm" thành vô lý và giải thích thêm)
\(\left(x+3\right)^4+\left(x+5\right)^4=2\)
<=> (x + 4 - 1)4 + (x + 4 + 1)4 - 2 = 0
Đặt y = x + 4
<=> (y - 1)4 + (y + 1)4 - 2 = 0
<=> y4 - 4y3 + 6y2 - 4y + 1 + y4 + 4y3 + 6y2 + 4y + 1 - 2 = 0
<=> 2y4 + 12y2 = 0
<=> 2y2(y2 + 6) = 0
<=> \(\orbr{\begin{cases}y^2=0\\y^2+6=0\left(ktm\right)\end{cases}}\)
<=> y = 0
<=> x + 4 = 0
<=> x = -4
Vậy S = {-4}
\(\frac{x^2+x+4}{2}+\frac{x^2+x+7}{3}=\frac{x^2+x+13}{5}+\frac{x^2+x+16}{6}\)
<=> \(\frac{x^2+x+4}{2}-3+\frac{x^2+x+7}{3}-3=\frac{x^2+x+13}{5}-3+\frac{x^2+x+16}{6}-3\)
<=> \(\frac{x^2+x+4-6}{2}+\frac{x^2+x+7-9}{3}=\frac{x^2+x+13-15}{5}+\frac{x^2+x+16-18}{6}\)
<=> \(\frac{x^2+x-2}{2}+\frac{x^2+x-2}{3}=\frac{x^2+x-2}{5}+\frac{x^2+x-2}{6}\)
<=> \(\left(x^2+2x-x-2\right)\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6}\right)=0\)
<=> (x + 2)(x - 1) = 0 (do \(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6}\ne0\))
<=> \(\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
Vậy S = {-2; 1}
câu cuối: + 3 vào sau các phân số của pt như trên
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
h.3x - 2/6 - 5 = 3 - 2(x + 7)/4
<=> 3x - 2 - 30/6 = 3 - 2(x + 7)/4
<=> 3x - 32/6 = 3 - 2x - 14/4
<=> 3x - 32/6 = -2x - 11/4
<=> 6x - 64/12 = -6x - 33/12
<=> 6x - 64 = -6x - 33 <=> 12x = 31 <=> x = 31/12
\(\left|x-\frac{2}{5}\right|+\frac{1}{2}=\frac{3}{4}\)
=> \(\left|x-\frac{2}{5}\right|=\frac{1}{4}\)
=> \(\orbr{\begin{cases}x-\frac{2}{5}=\frac{1}{4}\\x-\frac{2}{5}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{20}\\x=\frac{3}{20}\end{cases}}\)
b) \(\left|5-3x\right|+\frac{2}{3}=\frac{11}{6}\)
=> \(\left|5-3x\right|=\frac{7}{6}\)
=> \(\orbr{\begin{cases}5-3x=\frac{7}{6}\\5-3x=-\frac{7}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}3x=\frac{23}{6}\\3x=\frac{37}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{23}{18}\\x=\frac{37}{18}\end{cases}}\)
c) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=6^2\)
=> \(\left|x+\frac{3}{4}\right|=36,5\)
=> \(\orbr{\begin{cases}x+\frac{3}{4}=36,5\\x+\frac{3}{4}=-36,5\end{cases}}\Rightarrow\orbr{\begin{cases}x=35,75\\x=-37,25\end{cases}}\)
a) \(\left|x-\frac{2}{5}\right|+\frac{1}{2}=\frac{3}{4}\Leftrightarrow\left|x-\frac{2}{5}\right|=\frac{1}{4}\)\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{5}=\frac{1}{4}\\x-\frac{2}{5}=\frac{-1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{20}\\x=\frac{3}{20}\end{cases}}}\)
b) \(\left|5-3x\right|+\frac{2}{3}=\frac{11}{6}\Leftrightarrow\left|5-3x\right|=\frac{7}{6}\)\(\Leftrightarrow\orbr{\begin{cases}5-3x=\frac{7}{6}\\5-3x=\frac{-7}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=\frac{23}{6}\\3x=\frac{37}{6}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{23}{18}\\x=\frac{37}{18}\end{cases}}}\)
c) \(\frac{1}{5}-\left|\frac{1}{5}-x\right|=\frac{1}{5}\Leftrightarrow\left|\frac{1}{5}-x\right|=0\Leftrightarrow x=\frac{1}{5}\)
\(x+3\frac{1}{2}+x=24\frac{1}{4}\)
\(2x+\frac{7}{2}=\frac{97}{4}\)
\(2x=\frac{97}{4}-\frac{7}{2}\)
\(2x=\frac{83}{4}\)
\(x=\frac{83}{4}:2=\frac{83}{8}\)