\(\frac{4}{76}\)+\(\frac{3}{4}\)=??
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1/3x + 3/4x - 75% = -5 1/4
=> x - 3/4 = -21/4
=> x = -21/4 + 3/4
=> x = -9/2
60%x + 5/6x = -76
=> 3/5x + 5/6x = -76
=> 43/30x = -76
=> x = -76 : 43/30
=> x = -2280/43
\(\frac{1}{3}x+\frac{3}{4}x-75\%=-5\frac{1}{4}\)
\(\frac{13}{12}x-\frac{3}{4}=-\frac{21}{4}\)
\(\frac{13}{12}x=-\frac{9}{2}\)
\(x=-\frac{54}{13}\)
\(60\%.x+\frac{2}{3}x=-76\)
\(\frac{3}{5}x+\frac{2}{3}x=-76\)
\(\frac{19}{15}x=-76\)
\(x=-60\)
<=>\(\left(\frac{x+1}{77}+1\right)+\left(\frac{x+2}{76}+1\right)=\left(\frac{x+3}{75}+1\right)+\left(\frac{x+4}{74}+1\right)\)
<=> \(\frac{x+1+77}{77}+\frac{x+2+76}{76}=\frac{x+3+75}{75}+\frac{x+4+74}{74}\)
<=> \(\frac{x+78}{77}+\frac{x+78}{76}=\frac{x+78}{75}+\frac{x+78}{74}\)
<=> \(\frac{x+78}{77}+\frac{x+78}{76}-\frac{x+78}{75}-\frac{x+78}{74}\)
<=> \(\left(x+78\right)\left(\frac{1}{77}+\frac{1}{76}-\frac{1}{75}-\frac{1}{74}\right)\)
Vì \(\frac{1}{77}+\frac{1}{76}-\frac{1}{75}-\frac{1}{74}\ne0\) nên phương trình trên <=> x + 78 = 0
<=> x = -78
Tập nghiệm của phương trình trên là S= \(\left\{-78\right\}\)
Chúc bạn học tốt !
Ta có: \(\frac{x+1}{79}+\frac{x+4}{76}=-\frac{x+7}{73}-\frac{x+9}{71}-4\)
\(\Leftrightarrow\frac{x+1}{79}+\frac{x+4}{76}+\frac{x+7}{73}+\frac{x+9}{71}+4=0\)
\(\Leftrightarrow\frac{x+1}{79}+1+\frac{x+4}{76}+1+\frac{x+7}{73}+1+\frac{x+9}{71}+1=0\)
\(\Leftrightarrow\frac{x+80}{79}+\frac{x+80}{76}+\frac{x+80}{73}+\frac{x+80}{71}=0\)
\(\Leftrightarrow\left(x+80\right)\left(\frac{1}{79}+\frac{1}{76}+\frac{1}{73}+\frac{1}{71}\right)=0\)
Vì \(\frac{1}{79}+\frac{1}{76}+\frac{1}{73}+\frac{1}{71}>0\)
nên x+80=0
hay x=-80
Vậy: x=-80
\(a.\) \(3\frac{3}{4}+\left(4\frac{2}{4}-3\frac{1}{2}\right):\frac{3}{4}\)
\(=\frac{15}{4}\left(\frac{18}{4}-\frac{7}{2}\right):\frac{3}{4}\)
\(=\frac{15}{4}+\frac{4}{4}:\frac{3}{4}=\frac{15}{4}+\frac{4}{3}\)
\(=\frac{15}{4}+\frac{4}{3}=\frac{61}{12}\)
\(b.\) \(7\cdot\frac{2}{3}-\frac{2}{5}:\frac{1}{2}-\frac{2}{3}\)
\(=\frac{7}{3}-\frac{2}{5}\cdot2-\frac{2}{3}\)
\(=\frac{7}{3}-\frac{4}{5}-\frac{2}{3}=\frac{35-12-10}{15}\)
\(=\frac{13}{15}\)
\(c.\) \(68,7-100:20+70,8\)
\(=68,7-5+70,8\)
\(=63,7+70,8\)
\(=134,5\)
\(d.\)\(\left(5915+445:5\right)-76\cdot25\)
\(=\left(5915+89\right)-1900\)
\(=6004-1900=4104\)
a)=\(\frac{15}{4}+\left(\frac{18}{4}-\frac{7}{2}\right)\)/ \(\frac{3}{4}\)
=\(\left(\frac{33}{4}-\frac{14}{4}\right)\)/ \(\frac{3}{4}\)
= \(\frac{19}{4}\cdot\frac{4}{3}\)
=\(\frac{19}{3}\)
b) = \(\frac{2}{3}\cdot\left(7-1\right)-\frac{2}{5}\cdot\frac{2}{1}\)
= \(\frac{2}{3}\cdot6-\frac{4}{5}\)
= 1-\(\frac{4}{5}\)
= \(\frac{1}{5}\)
c) = 68.7 - 5 + 70.8
= 63.7 + 70.8
=134.5
d) = (5915 - 89) -76*25
= 5826 - 1900
= 3926
Bài làm
a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)
\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\frac{(3x+2)\left(3x+2\right)}{(3x-2)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{(3x+2)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)
\(\Leftrightarrow6x+4=0\)
\(\Leftrightarrow x=-\frac{4}{6}\)
\(\Leftrightarrow x=-\frac{2}{3}\)
Vậy x = -2/3 là nghiệm.
@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x2-4
Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)
x/2=y/5 ; y/3=z/4 ; z/6=t/11
<=> \(\frac{x}{6}=\frac{y}{15}=\frac{z}{20}\); z/6=t/11
<=> \(\frac{x}{36}=\frac{y}{90}=\frac{z}{120}=\frac{t}{220}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có
\(\frac{x}{36}=\frac{y}{90}=\frac{z}{120}=\frac{t}{220}=\frac{2x+y-z+\frac{t}{2}}{2.36+90-120+\frac{220}{2}}=\frac{-76}{152}=\frac{-1}{2}\)
Từ đó => ddc x,y,z