\(\dfrac{7}{5}X\dfrac{2}{5}\)
\(\dfrac{3}{4}+\dfrac{5}{4}:\dfrac{1}{4}X\dfrac{1}{2}\)
35864:175
247x328
ÉT O ÉT
6. ÉT O ÉT
\(\dfrac{9}{7}\) và \(\dfrac{5}{6}\) \(\dfrac{4}{7}\) và \(\dfrac{8}{21}\) \(\dfrac{2}{5};\dfrac{1}{3}\) và \(\dfrac{1}{2}\) \(\dfrac{1}{3};\dfrac{4}{3}\) và\(\dfrac{1}{5}\)
a: 9/7>1>5/6
b: 4/7=12/21>8/21
c: 2/5=12/30
1/3=10/30
1/2=15/30
mà 10<12<15
nên 1/3<2/5<1/2
98775 - 32 x 85 / 67500 - 24 x 236 / 568 + 101598 : 287 / 6875 + 980 -180 \(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\) / \(\dfrac{8}{11}+\dfrac{8}{33}\) x \(\dfrac{3}{4}\) / 7/9 x 3/14 :5/8 / \(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)
ÉT O ÉT
98775 - 32 x 85
=98775 -2720
=96055
67500 - 24 x 236
= 67500 -5664
=61836
568 + 101598 : 287
= 568 +354
=922
6875 + 980 -180
=7855 -180
=7675
\(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\)
\(=\dfrac{7}{10}-\dfrac{1}{2}\)
= \(\dfrac{1}{5}\)
\(\dfrac{8}{11}+\dfrac{8}{33}x\dfrac{3}{4}\)
\(=\dfrac{8}{11}+\dfrac{2}{11}\)
\(=\dfrac{10}{11}\)
\(\dfrac{7}{9}x\dfrac{3}{14}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}x\dfrac{8}{5}\)
\(=\dfrac{8}{30}\)
\(=\dfrac{4}{15}\)
\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)
\(=\dfrac{5}{12}-\dfrac{7}{32}x\dfrac{16}{21}\)
\(=\dfrac{5}{12}-\dfrac{1}{6}\)
\(=\dfrac{5}{12}-\dfrac{2}{12}\)
\(=\dfrac{3}{12}=\dfrac{1}{4}\)
Biết \(\dfrac{5z-3y}{2}\) = \(\dfrac{3x-2z}{5}\) = \(\dfrac{2y-5x}{3}\) Chứng minh: \(\dfrac{2}{x}\) = \(\dfrac{5}{y}\) = \(\dfrac{3}{z}\)
Ét ô ét! ☹
Áp dụng tính chất dãy tỉ số bằng nhau:\(\dfrac{\left(5z-3y\right)+\left(3x-2z\right)+\left(2y-5x\right)}{2+5+3}\)
=\(\dfrac{\left(3x-5x\right)+\left(-3y+2y\right)+\left(5z-2z\right)}{2+5+3}\)
=\(\dfrac{-2x-y+3z}{2+5+3}\)(???!!!!)
=\(\dfrac{-2x}{2}=\dfrac{-y}{5}=\dfrac{3z}{3}\)
=\(\dfrac{2}{-2x}=\dfrac{5}{-y}=\dfrac{3}{3z}\)
tớ xin chịu trận vì ko chứng minh được :(((
nó lại ra như thế này
ét ô ét mấy mắ uiii
cho hai đa thức sau : P(\(x\)) = 5\(x\)\(^5\)+3\(x\) - 4\(x\)\(^4\)- 2\(x\)\(^3\)+ 6 + 4\(x\)\(^2\)
Q(\(x\)) = 2\(x\)\(^4\)- x + 3x\(^{^{ }2}\)- 2x\(^3\)+\(\dfrac{1}{4}\)-x\(^5\)
a. Sắp xếp các hạng tử của mỗi đa thức theo lũy thừa giảm dần của biến ?
b. tính P(x) - Q(x)
c. chứng tỏ x=-1 là nghiệm của P(x) nhưng không là nghiệm của Q(x)
d. tính gtri của P(x) - Q(x) tại x=-1
\(a)P\left(x\right)=5x^5+3x-4x^4-2x^3+6+4x^2\)
\(P\left(x\right)=5x^5-4x^4-2x^3+4x^2+3x+6\)
\(Q\left(x\right)=2x^4-x+3x^2-2x^3+\dfrac{1}{4}-x^5\)
\(Q\left(x\right)=-x^5+2x^4-2x^3+3x^2-x+\dfrac{1}{4}\)
\(a)P\left(x\right)-Q\left(x\right)=\left(5x^5-4x^4-2x^3+4x^2+3x+6\right)+\left(-x^5+2x^4-2x^3+3x^2-x+\dfrac{1}{4}\right)\)
\(P\left(x\right)-Q\left(x\right)=5x^5-4x^4-2x^3+4x^2+3x+6+x^5-2x^4+2x^3-3x^2+x-\dfrac{1}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(5x^5+x^5\right)+\left(-4x^4-2x^4\right)+\left(-2x^3+2x^3\right)+\left(4x^2-3x^2\right)+\left(3x+x\right)+\left(6-\dfrac{1}{4}\right)\)
\(P\left(x\right)-Q\left(x\right)=6x^5-6x^4+x^2+4x+\dfrac{23}{4}\)
\(\text{c)Thay x=-1 vào biểu thức P(x),ta được:}\)
\(P\left(x\right)=5.\left(-1\right)^5-4.\left(-1\right)^4-2.\left(-1\right)^3+4.\left(-1\right)^2+3.\left(-1\right)+6\)
\(P\left(x\right)=\left(-5\right)-4-\left(-2\right)+4+\left(-3\right)+6\)
\(P\left(x\right)=\left(-9\right)-\left(-2\right)+4+\left(-3\right)+6\)
\(P\left(x\right)=\left(-7\right)+4+\left(-3\right)+6\)
\(P\left(x\right)=\left(-3\right)+\left(-3\right)+6\)
\(P\left(x\right)=\left(-6\right)+6=0\)
\(\text{Vậy giá trị của P(x) tại x=-1 là:0}\)
\(\text{Vậy =-1 là nghiệm của P(x)}\)
\(\text{Thay x=-1 vào biểu thức Q(x),ta được:}\)
\(Q\left(x\right)=\left(-1\right).5+2.\left(-1\right)^4-2.\left(-1\right)^3+3.\left(-1\right)^2-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-5\right)+2-\left(-2\right)+3-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-3\right)-\left(-2\right)+3-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-5\right)+3-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-2\right)-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-3\right)+\dfrac{1}{4}=\dfrac{-13}{4}\)
\(\text{Vậy x=-1 không phải là nghiệm của Q(x)}\)
\(\text{d)Thay x=-1 vào biểu thức }P\left(x\right)-Q\left(x\right),\text{ta được:}\)
\(P\left(x\right)-Q\left(x\right)=6.\left(-1\right)^5-6.\left(-1\right)^4+\left(-1\right)^2+4.\left(-1\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-6\right)-6+1+\left(-4\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-12\right)+1+\left(-4\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-11\right)+\left(-4\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-15\right)+\dfrac{23}{4}=\dfrac{-37}{4}\)
\(\text{Vậy giá trị của P(x)-Q(x) tại x=-1 là:}\dfrac{-37}{4}\)
k) 8 - \(\dfrac{x-2}{2}\) = \(\dfrac{x}{4}\)
m) \(\dfrac{3x+2}{2}\) - \(\dfrac{3x+1}{6}\) = 2x + \(\dfrac{5}{3}\)
n) \(\dfrac{x+1}{7}\)+ \(\dfrac{x+2}{6}\) = \(\dfrac{x+3}{5}\) + \(\dfrac{x+4}{4}\)
o) \(\dfrac{x+5}{6}\) + \(\dfrac{x+6}{5}\) = x + 9
\(\begin{array}{l} n) \Leftrightarrow \dfrac{{x + 1}}{7} + 1 + \dfrac{{x + 2}}{6} + 1 = \dfrac{{x + 3}}{5} + 1 + \dfrac{{x + 4}}{4} + 1\\ \Leftrightarrow \dfrac{{x + 8}}{7} + \dfrac{{x + 8}}{6} - \dfrac{{x + 8}}{5} - \dfrac{{x + 8}}{4} = 0\\ \Leftrightarrow \left( {x + 8} \right)\underbrace {\left( {\dfrac{1}{7} + \dfrac{1}{8} - \dfrac{1}{5} - \dfrac{1}{6}} \right)}_{ < 0} = 0\\ \Leftrightarrow x + 8 = 0\\ \Leftrightarrow x = - 8 \end{array}\)
k/
\(8-\dfrac{x-2}{3}=\dfrac{x}{4}\)
\(\Leftrightarrow\dfrac{96}{12}-\dfrac{4\left(x-2\right)}{12}=\dfrac{3x}{12}\)
\(\Leftrightarrow96-4x+8=3x\)
\(\Leftrightarrow96-4x+8-3x=0\)
\(\Leftrightarrow104-7x=0\)
\(\Leftrightarrow7x=104\)
\(\Leftrightarrow x=104:7\)
\(\Leftrightarrow x=\dfrac{104}{7}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{104}{7}\right\}\)
m/
\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow9x+6-3x-1-12x-10=0\)
\(\Leftrightarrow-6x-5=0\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{5}{6}\right\}\)
k) Ta có: \(8-\dfrac{x-2}{2}=\dfrac{x}{4}\)
\(\Leftrightarrow\dfrac{32}{4}-\dfrac{2\left(x-2\right)}{4}=\dfrac{x}{4}\)
\(\Leftrightarrow32-2x+4-x=0\)
\(\Leftrightarrow28-x=0\)
hay x=28
Vậy: S={28}
m) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
\(\Leftrightarrow6x+5-12x-10=0\)
\(\Leftrightarrow-6x=5\)
hay \(x=-\dfrac{5}{6}\)
Vậy: \(S=\left\{-\dfrac{5}{6}\right\}\)
n) Ta có: \(\dfrac{x+1}{7}+\dfrac{x+2}{6}=\dfrac{x+3}{5}+\dfrac{x+4}{4}\)
\(\Leftrightarrow\dfrac{x+1}{7}+1+\dfrac{x+2}{6}+1=\dfrac{x+3}{5}+1+\dfrac{x+4}{4}+1\)
\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}=\dfrac{x+8}{5}+\dfrac{x+8}{4}\)
\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}-\dfrac{x+8}{5}-\dfrac{x+8}{4}=0\)
\(\Leftrightarrow\left(x+8\right)\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)
mà \(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\ne0\)
nên x+8=0
hay x=-8
Vậy: S={-8}
Tìm x, biết:
a) \(\dfrac{-2}{5}\) + \(\dfrac{4}{5}\) . x = \(\dfrac{3}{5}\)
b) \(\dfrac{-3}{7}\) - \(\dfrac{4}{7}\) : x = \(\dfrac{2}{5}\)
c) \(\dfrac{4}{7}\) . x + \(\dfrac{2}{3}\) = \(\dfrac{-1}{5}\)
d) \(\dfrac{5}{7}\) : x -1 = \(\dfrac{2}{3}\)
a, - \(\dfrac{2}{5}\) + \(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)
\(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)+ \(\dfrac{2}{5}\)
\(\dfrac{4}{5}\).\(x\) = 1
\(x\) = \(\dfrac{5}{4}\)
b, - \(\dfrac{3}{7}\) - \(\dfrac{4}{7}\): \(x\) = \(\dfrac{2}{5}\)
\(\dfrac{4}{7}\): \(x\) = - \(\dfrac{3}{7}\) - \(\dfrac{2}{5}\)
\(\dfrac{4}{7}\): \(x\) = - \(\dfrac{29}{35}\)
\(x\) = \(\dfrac{4}{7}\): (- \(\dfrac{29}{35}\) )
\(x\) = - \(\dfrac{20}{29}\)
c, \(\dfrac{4}{7}\).\(x\) + \(\dfrac{2}{3}\) = - \(\dfrac{1}{5}\)
\(\dfrac{4}{7}\).\(x\) = -\(\dfrac{1}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{4}{7}\).\(x\) = - \(\dfrac{13}{15}\)
\(x\) = - \(\dfrac{13}{15}\): \(\dfrac{4}{7}\)
\(x\) = - \(\dfrac{91}{60}\)
d, \(\dfrac{5}{7}\): \(x\) - 1 = \(\dfrac{2}{3}\)
\(\dfrac{5}{7}\): \(x\) = \(\dfrac{2}{3}\)+ 1
\(\dfrac{5}{7}\): \(x\) = \(\dfrac{5}{3}\)
\(x\) = \(\dfrac{5}{7}\): \(\dfrac{5}{3}\)
\(x\) = \(\dfrac{3}{7}\)
Tìm 1 phân số \(\dfrac{x}{y}\) sao cho: \(\dfrac{7}{9}\)<\(\dfrac{x}{y}\)<\(\dfrac{7}{8}\) giải thích hộ mik luôn nha!
ét o ét
\(\dfrac{7}{9}< \dfrac{x}{y}< \dfrac{7}{8}\)
\(\dfrac{56}{72}< \dfrac{x}{y}< \dfrac{63}{72}\)
\(\dfrac{x}{y}=\left\{\dfrac{57}{72};\dfrac{58}{72};\dfrac{59}{72};\dfrac{60}{72};\dfrac{61}{72};\dfrac{62}{72}\right\}\)
a,\(\dfrac{2}{3}\)x\(\dfrac{5}{2}\):\(\dfrac{9}{5}\)
b,\(\dfrac{1}{3}\)x\(\dfrac{1}{4}\)+\(\dfrac{5}{6}\)
c,\(\dfrac{1}{2}\)-\(\dfrac{7}{8}\):\(\dfrac{7}{4}\)
d,\(\dfrac{6}{5}\)-\(\dfrac{4}{5}\)x\(\dfrac{3}{2}\)
a) \(2\dfrac{1}{2}\) - x + \(\dfrac{4}{5}\) = \(\dfrac{2}{3}\) - (\(-\dfrac{4}{7}\) )
b) \(-\dfrac{4}{7}\) - x = \(\dfrac{3}{5}\) - 2x
c) (\(\dfrac{3}{8}\) - \(\dfrac{1}{5}\) ) + (\(\dfrac{5}{8}\) - x) = \(\dfrac{1}{5}\)
\(a,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \Rightarrow\dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{26}{21}\\ \Rightarrow\dfrac{5}{2}-x=\dfrac{46}{105}\\ \Rightarrow x=\dfrac{433}{210}\\ b,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow2x-\dfrac{4}{7}-x=\dfrac{3}{5}\\ \Rightarrow2x-x=\dfrac{41}{35}\\ \Rightarrow x=\dfrac{41}{35}\\ c,\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{40}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{5}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{3}{5}.\)