10,5-(x+ 1 phần 3)=3 phần 2
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AK
9 tháng 4 2018
a ) \(-\frac{3}{7}.\frac{3}{11}+-\frac{3}{7}.\frac{8}{11}+1\frac{3}{7}\)
\(=-\frac{3}{7}.\left(\frac{3}{11}+\frac{8}{11}\right)+\frac{10}{7}\)
\(=-\frac{3}{7}.\frac{11}{11}+\frac{10}{7}\)
\(=-\frac{3}{7}.1+\frac{10}{7}\)
\(=\frac{10}{7}\)
b ) \(75\%.10,5=\frac{3}{4}.10,5=7,875\)
c ) \(5-3.\left(\left|-4\right|-30:15\right)\)
\(=5-3.\left(4-2\right)\)
\(=5-3.2\)
\(=5-6\)
\(=-1\)
d ) \(-\frac{5}{7}.\frac{2}{11}+-\frac{5}{7}.\frac{9}{11}+1\frac{5}{7}\)
\(=-\frac{5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)
\(=-\frac{5}{7}.1+\frac{12}{7}\)
\(=\frac{7}{7}\)
\(=1\)
Chúc bạn học tốt !!!
PN
6 tháng 8 2020
\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)
\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)
\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)
\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)
\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)
\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)
\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)
\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)
\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)
\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)
\(< =>5x-35=32x-8< =>32x-5x=-35+8\)
\(< =>27x=-27< =>x=-1\)
PN
6 tháng 8 2020
\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)
\(< =>12=6x-3< =>6x=12+3\)
\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)
\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)
\(< =>8x-4=9x+3< =>9x-8x=-4-3\)
\(< =>9x-8x=-7< =>x=-7\)
\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)
\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)
\(< =>12x-7x=14-4< =>5x=10\)
\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)
\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)
\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)
\(< =>6x-4x=4+6< =>2x=10\)
\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)
\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)
\(< =>\left(x+1\right)\left(x+1\right)=3.3\)
\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)
\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)
\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)
\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)
AT
26 tháng 4 2018
heoheo lần sau bạn đánh = kí hiệu đi :(((
a/ \(\dfrac{x}{3}+\dfrac{2x-1}{6}=\dfrac{1}{2}\)
\(\Leftrightarrow2x+2x-1=3\)
<=> 4x = 4 <=> x = 1
Vậy x = 1
b/ \(\dfrac{3x+1}{2}+\dfrac{x-1}{3}=\dfrac{x-9}{6}\)
\(\Leftrightarrow3\left(3x+1\right)+2\left(x-1\right)=x-9\)
\(\Leftrightarrow9x+3+2x-2=x-9\)
\(\Leftrightarrow10x=-10\Leftrightarrow x=-1\)
Vậy pt có nghiệm x = -1
c/ \(\dfrac{x-1}{x-2}=\dfrac{x+3}{x+2}\) ĐKXĐ: \(x\ne\pm2\)
<=> \(\left(x-1\right)\left(x+2\right)=\left(x+3\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+2x-x-2=x^2-2x+3x-6\)
\(\Leftrightarrow0x=-4\left(voly\right)\)
Vậy pt vô nghiệm
d/ \(\dfrac{3x-1}{3x+1}+\dfrac{x-3}{x+3}=2\) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-3\\x\ne-\dfrac{1}{3}\end{matrix}\right.\)
pt <=> \(\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(3x+1\right)\left(x+3\right)}+\dfrac{\left(x-3\right)\left(3x+1\right)}{\left(3x+1\right)\left(x+3\right)}=\dfrac{2\left(3x+1\right)\left(x+3\right)}{\left(3x+1\right)\left(x+3\right)}\)
=> (3x-1)(x+3) + (x-3)(3x+1) = 2(3x+1)(x+3)
\(\Leftrightarrow3x^2+8x-3+3x^2-8x-3=6x^2+20x+6\)
\(\Leftrightarrow-20x=12\Leftrightarrow x=-\dfrac{3}{5}\left(tm\right)\)
Vậy pt có nghiệm x=....
e/ như ý d
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 4
(\(x\) + \(\dfrac{1}{2}\))2 = \(\dfrac{1}{16}\)
\(\left[{}\begin{matrix}x+\dfrac{1}{2}=-\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{4}-\dfrac{1}{2}\\x=\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {- \(\dfrac{3}{4};-\dfrac{1}{4}\)}
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 4
\(x\) : (- \(\dfrac{1}{3}\))3 = - \(\dfrac{1}{3}\)
\(x\) = (-\(\dfrac{1}{3}\)).(-\(\dfrac{1}{3}\))3
\(x\) = \(\dfrac{1}{81}\)
Vậy \(x=\dfrac{1}{81}\)
Đề bài: Tìm x
\(10,5-\left(x+\frac{1}{3}\right)=\frac{3}{2}\)
\(\frac{105}{10}-\left(x+\frac{1}{3}\right)=\frac{3}{2}\)
\(\frac{21}{2}-\left(x+\frac{1}{3}\right)=\frac{3}{2}\)
\(x+\frac{1}{3}=\frac{21}{2}-\frac{3}{2}\)
\(x+\frac{1}{3}=\frac{18}{2}\)
\(x+\frac{1}{3}=9\)
\(x=9-\frac{1}{3}\)
\(x=\frac{27}{3}-\frac{1}{3}\)
\(x=\frac{26}{3}\)
Vậy \(x=\frac{26}{3}\).