\(\dfrac{1}{3}+\dfrac{2}{3}.x=\dfrac{4}{3}\)
\(\dfrac{3}{5}.x+\dfrac{1}{4}=\dfrac{1}{10}\)
1/ \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
2/ \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
3/ \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
4/ \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
5/ \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)
\(\Leftrightarrow5x+1-2x+4=4\)
\(\Leftrightarrow3x=-1\)
hay \(x=-\dfrac{1}{3}\)
2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
\(\Leftrightarrow9x+27+12-36x=-2x+2\)
\(\Leftrightarrow-27x+2x=2-39\)
hay \(x=\dfrac{37}{25}\)
3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x+6-10x=4-4x\)
\(\Leftrightarrow-7x+4x=4-6=-2\)
hay \(x=\dfrac{2}{3}\)
4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
\(\Leftrightarrow5x-15-x-1=2x-4\)
\(\Leftrightarrow4x-2x=-4+16=12\)
hay x=6
5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
\(\Leftrightarrow12x+3-9x+5+4x-8=0\)
\(\Leftrightarrow7x=0\)
hay x=0
Tìm x biết:
\(a,\dfrac{4}{5}+x=\dfrac{2}{3}\)
\(b,\dfrac{-5}{6}-x=\dfrac{2}{3}\)
\(c,\dfrac{1}{2}x+\dfrac{3}{4}=\dfrac{-3}{10}\)
\(d,\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\)
\(e,\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(h,x+30\%x=-1,3\)
\(k,3\dfrac{1}{3}x+16\dfrac{1}{4}=13,25\)
\(m,\dfrac{x-6}{2}=\dfrac{50}{x-6}\)
\(n,x-13,4=24,5-6,7.5,2\)
\(p,15,7x+3,6x=-96,5\)
\(q,2,5x-11,6=-59,1\)
a)4/5+x=2/3
x=2/3-4/5
x=-2/15
b)-5/6-x=2/3
x=-5/6-2/3
x=-3/2
c)1/2x+3/4=-3/10
1/2x=-3/10-3/4
1/2x=-21/20
x=-21/20:1/2
x=-21/10
d)x/3-1/2=1/5
x/3=1/5+1/2
x/3=7/10
10x/30=21/30
10x=21
x=21:10
x=21/10
giải các phương trinh sau
1/ \(\dfrac{4x-4}{3}-\dfrac{7-x}{5}\)
2/ \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
3/ \(\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\)
4/ \(\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\)
5/ \(\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\)
\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
Bài 3 Tìm x
\(\dfrac{1}{2}:3+x=1\dfrac{2}{3}\)
\(2\dfrac{3}{4}-x=\dfrac{5}{6}+\dfrac{2}{3}\)
\(5\dfrac{4}{10}-x\) X\(\dfrac{3}{4}=\dfrac{2}{3}\)
\(\dfrac{1}{2}:3+x=1\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{5}{3}-\dfrac{1}{6}\\ \Leftrightarrow x=\dfrac{3}{2}\\2\dfrac{3}{4}-x=\dfrac{5}{6}+\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{11}{4}-\dfrac{5}{6}-\dfrac{2}{3} \\ \Leftrightarrow x=\dfrac{5}{4}\\ 5\dfrac{4}{10}-\dfrac{3}{4}\times x=\dfrac{2}{3}\\ \Leftrightarrow\dfrac{3}{4}x=\dfrac{54}{10}-\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{284}{45}\)
1) ....
1/2 : 3 = 5/3 - x
1/6 = 5/3 - x
x = 5/3 - 1/6 =3/2
2)....
11/4 - x = 3/2
x = 11/4 - 3/2 =5/4
3)...
27/5 - 3/4x = 2/3
3/4x = 27/5 - 2/3 =71/15
x = 71/15 : 3/4 =284/45
câu 1: (x+\(\dfrac{1}{2}\)).(\(\dfrac{2}{3}\)-2x)=0
câu 2: (3x-10)(-\(\dfrac{1}{2}\)x+5)=0
câu 3: \(\dfrac{1}{3}\)x+\(\dfrac{53}{4}\)=\(\dfrac{65}{4}\)
câu 4: \(\dfrac{2}{3}\)x-\(\dfrac{4}{9}\)=\(\dfrac{2}{9}\)
câu 5: \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{x\left(x+1\right)}\)=\(\dfrac{2010}{2011}\)
Câu 1:
\(\Rightarrow \left[\begin{array}{} x+\frac{1}{2}=0\\ \frac{2}{3}-2x=0 \end{array} \right.\)
\(\Leftrightarrow \left[\begin{array}{} x=\frac{-1}{2}\\ x=\frac{1}{3} \end{array} \right.\)
Vậy phương trình có tập nghiệm S={\(\frac{-1}{2};\frac{1}{3}\)}
Câu 2:
\(\Rightarrow \left[\begin{array}{} 3x-10=0\\ 5-\frac{1}{2}x=0 \end{array} \right.\)
\(\Leftrightarrow \left[\begin{array}{} x-=\frac{10}{3}\\ x=10 \end{array} \right.\)
Vậy phương trình có tập nghiệm S={\(10;\frac{10}{3}\)}
Câu 3:
\(\Leftrightarrow \frac{1}{3}x=\frac{65}{4}-\frac{53}{4}\)
\( \Leftrightarrow \frac{1}{3}x=\frac{12}{4}\)
\(\Leftrightarrow x=9\)
Vậy phương trình có tập nghiệm S={9}
Câu 4:
\(\Leftrightarrow \frac{2}{3}x=\frac{2}{3}\)
\(\Leftrightarrow x=1\)
Vậy phương trình có tập nghiệm S={1}
Câu 5:
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x(x+1)}=\frac{2010}{2011}\)
\(\Leftrightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)
\(\Leftrightarrow 1-\frac{1}{x+1}=\frac{2010}{2011}\)
\(\Leftrightarrow \frac{x}{x+1}=\frac{2010}{2011}\)
\(\Rightarrow 2010x+2010=2011x\)
\(\Leftrightarrow x=2010\)
Vậy phương trình có tập nghiệm S={2010}
cảm ơn bạn Hoàng Bình Bảo nha nhưng mà đây là toán lớp 6 mà bạn
Bài 2: (đề 2) Tìm y
a) \(2\dfrac{2}{5}-y:2\dfrac{3}{4}=1\dfrac{1}{2}\) b) \(1\dfrac{1}{4}+2\dfrac{1}{5}\) x \(y=2\dfrac{3}{5}\)
c) \(2\dfrac{4}{5}-2\dfrac{1}{4}:y=\dfrac{3}{4}\) c) \(x:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\)
\(2\dfrac{2}{5}-y:2\dfrac{3}{4}=1\dfrac{1}{2}\\ \dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{9}{10}\\ y=\dfrac{9}{10}\times\dfrac{11}{4}=\dfrac{99}{40}\\ b,1\dfrac{1}{4}+2\dfrac{1}{5}\times y=2\dfrac{3}{5}\\ \dfrac{5}{4}+\dfrac{11}{5}\times y=\dfrac{13}{5}\\ \dfrac{11}{5}\times y=\dfrac{13}{5}-\dfrac{5}{4}\\ \dfrac{11}{5}\times y=\dfrac{27}{20}\\ y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{44}\)
\(c,2\dfrac{4}{5}-2\dfrac{1}{4}:y=\dfrac{3}{4}\\ \dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{41}{20}\\ y=\dfrac{9}{4}:\dfrac{41}{20}=\dfrac{45}{41}\\ c2,x:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{31}{10}\\ x=\dfrac{31}{10}\times\dfrac{10}{3}=\dfrac{31}{3}\)
a) \(...\Rightarrow\dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\)
\(\Rightarrow y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\Rightarrow y:\dfrac{11}{4}=\dfrac{24}{10}-\dfrac{15}{10}\)
\(\Rightarrow y:\dfrac{11}{4}=\dfrac{9}{10}\Rightarrow y=\dfrac{9}{10}x\dfrac{11}{4}=\dfrac{99}{40}\)
b) \(...\Rightarrow\dfrac{5}{4}+\dfrac{11}{5}xy=\dfrac{13}{5}\Rightarrow\dfrac{11}{5}xy=\dfrac{13}{5}-\dfrac{5}{4}\)
\(\Rightarrow\dfrac{11}{5}xy=\dfrac{52}{20}-\dfrac{25}{20}\Rightarrow\dfrac{11}{5}xy=\dfrac{27}{20}\)
\(\Rightarrow y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{20}x\dfrac{5}{11}=\dfrac{27}{44}\)
c) \(...\Rightarrow\dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\Rightarrow\dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{9}{4}:y=\dfrac{56}{20}-\dfrac{15}{20}\Rightarrow\dfrac{9}{4}:y=\dfrac{39}{20}\)
\(\Rightarrow y=\dfrac{9}{4}:\dfrac{39}{20}\Rightarrow y=\dfrac{9}{4}x\dfrac{20}{39}=\dfrac{15}{13}\)
d) \(...\Rightarrow x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\Rightarrow x:\dfrac{10}{3}=\dfrac{24}{10}+\dfrac{7}{10}\)
\(\Rightarrow x:\dfrac{10}{3}=\dfrac{31}{10}\Rightarrow x=\dfrac{31}{10}x\dfrac{10}{3}=\dfrac{31}{3}\)
Tính :
\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)
\(\dfrac{1}{5}\) : \(4\) + \(\dfrac{3}{4}\)
\(\dfrac{4}{3}\) x \(\dfrac{9}{5}\) - \(\dfrac{3}{10}\)
\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{3+2}{6}-\dfrac{1}{5}=\dfrac{5}{6}-\dfrac{1}{5}=\dfrac{5\times5-6}{30}=\dfrac{19}{30}\\ \dfrac{1}{5}:4+\dfrac{3}{4}=\dfrac{1}{5}\times4+\dfrac{3}{4}=\dfrac{4}{5}+\dfrac{3}{4}=\dfrac{4\times4+3\times5}{20}=\dfrac{31}{20}\)
\(\dfrac{4}{3}\times\dfrac{9}{5}-\dfrac{3}{10}=\dfrac{36}{15}-\dfrac{3}{10}=\dfrac{12}{5}-\dfrac{3}{10}=\dfrac{12\times2-3}{10}=\dfrac{21}{10}\)
\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)
= \(\dfrac{1\times15}{2\times15}\) + \(\dfrac{1\times10}{3\times10}\) - \(\dfrac{1\times6}{5\times6}\)
= \(\dfrac{15}{30}\) + \(\dfrac{10}{30}\) - \(\dfrac{6}{30}\)
= \(\dfrac{19}{30}\)
\(\dfrac{1}{5}\) : 4 + \(\dfrac{3}{4}\)
= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)
= \(\dfrac{1}{20}\) + \(\dfrac{3\times5}{4\times5}\)
= \(\dfrac{1}{20}+\dfrac{15}{20}\)
= \(\dfrac{16}{20}\)
= \(\dfrac{4}{5}\)
\(\dfrac{4}{3}\) \(\times\) \(\dfrac{9}{5}\) - \(\dfrac{3}{10}\)
= \(\dfrac{12}{5}\) - \(\dfrac{3}{10}\)
= \(\dfrac{24}{10}\) - \(\dfrac{3}{10}\)
= \(\dfrac{21}{10}\)
Bài 2 Tìm x
a) 4\(\dfrac{2}{3}:x-\dfrac{3}{5}=\dfrac{4}{3}\)
b) \(\dfrac{4}{3}+2\dfrac{1}{2}\)x \(X=2\)
c) \(X:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\)
a) ....
=> 14/3 : x = 4/3 + 3/5 = 29/15
=> x = 14/3 : 29/15 = 70/29
b) ...
=> 4/3 + 5/2x = 2
=> 5/2x = 2 - 4/3 = 2/3
=> x = 2/3 : 5/2 = 4/15
c)...
=> x : 10/3 = 31/10
=> x = 31/10 x 10/3 = 31/3
tìm x biết
a , \(\dfrac{-3}{10}\)-(\(\dfrac{-1}{5}\) +x)=\(\dfrac{-3}{2}\)
b, -(-x+ \(\dfrac{3}{4}\))- \(\dfrac{12}{8}\).- \(\dfrac{32}{15}\)=- \(\dfrac{-1}{2}\)
c, \(\dfrac{x-3}{x+5}\)=\(\dfrac{4}{3}\)
Giải:
a)-3/10-(-1/5)+x)=-3/2
-1/5+x =-3/10-(-3/2)
-1/5+x =6/5
x =6/5-(-1/5)
x =7/5
b)-(-x+3/4)-12/8.(-32/15)=-(-1/2)
x-3/4+16/5 =1/2
x-3/4 =1/2-16/5
x =-27/10
x =-27/10+3/4
x =-39/20
c)x-3/x+5=4/3
=>(x-3).3=4.(x+5)
3x-9 =4x+20
3x-4x =20+9
-1x =29
x =-29
Câu b cậu nên tính lại cho kĩ nhé, ấn máy tính dễ nhầm lắm đấy!
Mk phải ấn: -(39/20+3/4)-12/8.-32/15=1/2
Vì x là số âm mà đằng trước x là dấu ''-'' nên -(-39/20)=39/20 ; -(-1/2)=1/2
Chúc bạn học tốt!