Tìm x:
\(2\frac{3}{4}x=3\frac{1}{7}:0,01\)
\(2\frac{1}{3}:\frac{1}{3}=\frac{7}{9}:x\)
\(3,2x+\left(-1,2\right)x+2,7=-4,9\)
\(x:\frac{9}{14}=\frac{7}{3}:x\)
\(\frac{37-x}{x+13}=\frac{3}{7}\)
\(\frac{x}{15}=\frac{-60}{x}\)
a) 3,2x + (-1,2)x + 2,7 = -4,9
b) \(\left(\frac{2}{3}x-1\right)\) \(\left(\frac{3}{4}x+\frac{1}{2}\right)\) + 0
c) x : \(\frac{9}{14}=\frac{7}{3}\) : x
d) ( 3x - 1 )10 = ( 3x-1)20
GIÚP MIH. CÁC BAJ CỐ GẮNG LÀM ĐƯỢC CÂU NÀO THÌ LÀM GIÚP MIH NHA. MIH CẢM ONQ
b) \(\left(\frac{2}{3}x-1\right).\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-1=0\\\frac{3}{4}x+\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=1\\\frac{3}{4}x=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1:\frac{2}{3}\\x=\left(-\frac{1}{2}\right):\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}.\)
c) \(x:\frac{9}{14}=\frac{7}{3}:x\)
\(\Rightarrow\frac{x}{\frac{19}{4}}=\frac{\frac{7}{3}}{x}\)
\(\Rightarrow x.x=\frac{7}{3}.\frac{19}{4}\)
\(\Rightarrow x.x=\frac{133}{12}\)
\(\Rightarrow x^2=\frac{133}{12}\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\frac{133}{12}}\\x=-\sqrt{\frac{133}{12}}\end{matrix}\right.\)
Vậy \(x\in\left\{\sqrt{\frac{133}{12}};-\sqrt{\frac{133}{12}}\right\}.\)
d) \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)
\(\Rightarrow\left(3x-1\right)^{10}-\left(3x-1\right)^{20}=0\)
\(\Rightarrow\left(3x-1\right)^{10}.\left[1-\left(3x-1\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^{10}=0\\1-\left(3x-1\right)^{10}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-1=0\\\left(3x-1\right)^{10}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x-1=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1:3\\3x-1=1\\3x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\3x=2\\3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=\frac{2}{3}\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{3};\frac{2}{3};0\right\}.\)
Chúc bạn học tốt!
Tìm x:
a, x + 30%x = -1,31
b,\(\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)
c, \(\frac{1}{2}x-\frac{3}{4}=\frac{14}{9}.\frac{3}{7}\)
d,\(\frac{-5}{6}-x=\frac{7}{12}+\frac{-1}{3}\)
e,\(\frac{x+3}{-15}=\frac{1}{3}\)
f,\(\left(4,5-2x\right).\left(-1\frac{4}{7}\right)=\frac{11}{14}\)
GIÚP MIK LM TRONG TỐI NAY NHA, MAI MIK PHẢI NỘP RỒI
\(a)x+30\%x=-1,31\)
\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)
\(\Leftrightarrow10x+3x=-13,1\)
\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)
\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)
\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=9\)
\(\Leftrightarrow6x-3=18\)
\(\Leftrightarrow x=\frac{7}{2}\)
\(c)\frac{1}{2}x-\frac{3}{4}=\frac{14}{9}.\frac{3}{7}\)
\(\Leftrightarrow\frac{x}{2}-\frac{3}{4}=\frac{2}{3}\)
\(\Leftrightarrow\frac{x}{2}=\frac{17}{12}\)
\(\Leftrightarrow12x=34\Leftrightarrow x=\frac{17}{6}\)
bài 1 : tìm x biết
a, \(\frac{2}{3}\times\left(x-\frac{5}{6}\right)+\frac{1}{4}=\frac{22}{9}\)
b, \(\frac{2}{3}:\frac{x}{5}=\frac{10}{21}\)
c, \(\frac{7}{3}:\frac{x}{5}=\frac{14}{15}\)
d, \(1-\left(5\frac{4}{9}\times x-7\frac{7}{18}\right):15\frac{3}{4}=0\)
bài 2 : tính gtri bt
a,\(\frac{8}{7}+\frac{1}{5}\times\frac{10}{9}\)
b, \(\frac{3}{2}+\left(\frac{9}{2}+\frac{2}{9}\right)\times\left(\frac{4}{3}-\frac{5}{4}\right)\)
!_ove
a) x = 99/20
b) x = 7
c) x = 2
( chỉ lm đc đến đó thui nk )
bài 2 tìm x
a,\(\frac{-2}{3}.x+\frac{1}{5}=\frac{3}{10}\)
b,\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
c,\(\frac{2}{3}.x-\frac{1}{2}=\frac{1}{10}\)
d,\(\frac{3}{5}+\frac{4}{9}:x=\frac{2}{3}\)
e,\(\left|x+75\%\right|=2\frac{1}{5}\)
i,\(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2.x\right)=0\)
k,\(\frac{4}{7}.x-\frac{2}{3}=\frac{1}{5}\)
l,\(\frac{2}{3}.x-\frac{3}{2}.x=\frac{5}{12}\)
m,\(\left|2.x-\frac{1}{3}\right|+\frac{5}{6}=1\)
n,\(\frac{1}{3}-\frac{7}{8}.x=\frac{1}{3}\)
11,\(\frac{x+2}{5}=\frac{7}{12}-1\frac{1}{4}\)
12,\(\left(2\frac{4}{5}.x-50\right):\frac{2}{3}=51\)
13,\(\frac{2}{5}+\frac{3}{5}.\left(3.x-3,7\right)=-\frac{53}{10}\)
14,\(\frac{7}{9}:\left(2+\frac{3}{4}.x\right)+\frac{5}{9}=\frac{23}{27}\)
Tìm x:
\(\frac{\left(13\frac{2}{9}-15\frac{2}{3}\right)\cdot\left(30^2-5^4\right)}{\left(18\frac{3}{7}-17\frac{1}{4}\right)\cdot\left(25-12\cdot5^2\right)}\cdot x=\frac{\frac{2}{11}+\frac{3}{13}+\frac{4}{15}+\frac{5}{17}}{4\frac{1}{11}+\frac{5}{13}+\frac{9}{15}+\frac{13}{17}}\)
Tìm x, biết:
a)\(x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\)
b)\(\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3};\)
c)\(\frac{2}{5}:x = \frac{1}{{16}}:0,125\)
d)\( - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\)
a)
\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)
Vậy \(x = \frac{{ - 3}}{2}\).
b)
\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)
Vậy \(x = \frac{{ - 5}}{6}\).
c)
\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)
Vậy \(x = \frac{4}{5}\)
d)
\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)
Vậy \(x = \frac{{ - 2}}{5}\).
Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.
Bai 1: a) \(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\)
b) \(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
bài 2:
a) \(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\)
b) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
c)
\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)
\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)
\(< =>\frac{-15+12-24}{63}\)
\(< =>\frac{-3}{7}\)
\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{7}\)
\(< =>\frac{29}{35}\)
\(bai2:\)
\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)
\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)
\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)
\(< =>x=\frac{3}{5}:\frac{-3}{4}\)
\(< =>x=\frac{-4}{5}\)
\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)
\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)
Bài 1:
a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\)
\(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\)
\(=\frac{29}{35}\)
Bài 2:
a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)
\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)
\(x.\frac{10}{3}=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{10}{3}\)
\(x=\frac{13}{40}\)
Tìm điều kiện xác định rồi giải các phương trình sau:
a) \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
b) \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
c) \(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
Help me!
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
Tìm x biết : \(\frac{\frac{2}{3}+\frac{1}{3}\left(1,2-x\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}=\frac{1}{4}\)