\(\left(5^4+5^7\right)\cdot\left(8^9-2^7\right)\cdot\left(2^4-4^2\right)\)
\(\dfrac{\left(4\times7+2\right)\left(6\times6+2\right)\left(8\times11+2\right)\cdot\cdot\cdot\left(100\times103+2\right)}{\left(5\times8+2\right)\left(7\times10+2\right)\left(9\times12+2\right)\cdot\cdot\cdot\left(99\times102+2\right)}=...\)
\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)
\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0.5\cdot\left(-2\dfrac{3}{5}\right)\)
\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)
\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)
\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)
\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)
\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)
Tính giá trị các biểu thức sau theo cách hợp lí nhất.
a) $\mathrm{A}=\left(\dfrac{2}{7} \cdot \dfrac{1}{4}-\dfrac{1}{3} \cdot \dfrac{2}{7}\right):\left(\dfrac{2}{7} \cdot \dfrac{3}{9}-\dfrac{2}{7} \cdot \dfrac{2}{5}\right)$;
b) $\mathrm{B}=\dfrac{\left(\dfrac{1}{5}-\dfrac{2}{7}\right) \cdot \dfrac{3}{4}-\dfrac{3}{4} \cdot\left(\dfrac{1}{3}-\dfrac{2}{7}\right)}{\dfrac{1}{5} \cdot \dfrac{2}{7}-\dfrac{1}{3} \cdot\left(\dfrac{2}{7}+\dfrac{3}{9}\right)+\dfrac{3}{9} \cdot \dfrac{1}{5}} .$
a) \(\dfrac{\left(x+\dfrac{3}{4}\right)\cdot\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
b)\(\dfrac{\left(5-\dfrac{2}{7}\right)\cdot\dfrac{7}{9}\cdot\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
Giải:
a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)
\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)
\(\Leftrightarrow x=-\dfrac{79}{210}\)
Vậy \(x=-\dfrac{79}{210}\).
b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)
\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)
\(\Leftrightarrow3x=\dfrac{67}{75}\)
\(\Leftrightarrow x=\dfrac{67}{75}:3\)
\(\Leftrightarrow x=\dfrac{67}{225}\)
Vậy \(x=\dfrac{67}{225}\).
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tình hợp lí:
\(\frac{\left(\frac{1}{5}-\frac{2}{7}\right)\cdot\frac{3}{4}-\frac{3}{4}\cdot\left(\frac{1}{3}-\frac{2}{7}\right)}{\frac{1}{5}\cdot\frac{2}{7}-\frac{1}{3}\cdot\left(\frac{2}{7}+\frac{3}{9}\right)+\frac{3}{9}+\frac{1}{5}}\)
Tính giá trị biểu thức :
1. \(A=\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}\)
2. \(B=\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
3. \(C=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4\cdot6}\cdot\dfrac{5^2}{4\cdot6}\)
4. \(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right)\cdot\left(\dfrac{2}{3}\cdot\dfrac{1}{4}\right)^2\)
5. Cho \(M=8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\) ; \(N=\left(10\dfrac{2}{9}+2\dfrac{3}{5}\right)-6\dfrac{2}{9}\). Tính \(P=M-N\)
6. \(E=10101\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3\cdot7\cdot11\cdot13\cdot37}\right)\)
7. \(F=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{256}+\dfrac{3}{64}}{1-\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
8. \(G=\text{[}\dfrac{\left(6-4\dfrac{1}{2}\right):0,03}{\left(3\dfrac{1}{20}-2,65\right)\cdot4+\dfrac{2}{5}}-\dfrac{\left(0,3-\dfrac{3}{20}\right)\cdot1\dfrac{1}{2}}{\left(1,88+2\dfrac{3}{25}\right)\cdot\dfrac{1}{80}}\text{]}:\dfrac{49}{60}\)
9. \(H=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{4\cdot5\cdot6}+...+\dfrac{1}{98\cdot99\cdot100}\)
10. \(I=\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}\)
11. \(K=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{999}\right)\)
12. \(L=1\dfrac{1}{3}+1\dfrac{1}{8}+1\dfrac{1}{15}...\) (98 thừa số)
13. \(M=-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{3}}}}\)
14. \(N=\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}\)
15. \(P=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{5}-1\right)...\left(\dfrac{1}{2001}-1\right)\)
16. \(Q=\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2005\cdot2006}\right):\left(\dfrac{1}{1004\cdot2006}+\dfrac{1}{1005\cdot2005}+...+\dfrac{1}{2006\cdot1004}\right)\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
1:tìm x
a; \(3x+\left|x-2\right|=8\)
b; \(5-\left|x-1\right|=4\)
2:tìm x
\(5\cdot\left(x-2\right)-4\cdot\left(1-3x\right)=\left|3-7\right|+2\cdot\left(1+2x\right)\)
3: tìm x
\(\left(x-2\right)\cdot\left(2x+1\right)-3\cdot\left(x+2\right)=4-5\cdot\left(1-x\right)\)
4:tìm x
\(1\dfrac{1}{2}\cdot\left(x-2\right)-\dfrac{x-5}{3}=3\dfrac{1}{3}\cdot\left(1-2x\right)-\dfrac{5\cdot\left(x+1\right)}{6}\)
5: tìm x
\(\left(x-3\right)\cdot\left(1-x\right)+\left(x-2\right)^2=\left(1-x\right)^2-2\cdot\left(1+x\right)\)
6: tìm x
\(\left(2x-1\right)^2-3\cdot\left(x+2\right)^2=4\cdot\left(x-2\right)-5\cdot\left(x-1\right)^2\)
1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ
2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ
4. 1\(\dfrac{1}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = 3\(\dfrac{1}{3}\).(1 - 2x) - \(\dfrac{5.\left(x+1\right)}{6}\)
<=> \(\dfrac{3}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = \(\dfrac{10}{3}\).(1 - 2x) - \(\dfrac{5x+5}{6}\)
<=> \(\dfrac{3}{2}x-3-\dfrac{x}{3}+\dfrac{5}{3}=\dfrac{10}{3}-\dfrac{20}{3}x-\dfrac{5x}{6}-\dfrac{5}{6}\)
<=> \(\dfrac{3}{2}x-\dfrac{x}{3}+\dfrac{20}{3}x-\dfrac{5x}{6}=\dfrac{10}{3}-\dfrac{5}{6}-3+\dfrac{5}{3}\)
<=> 7x = \(\dfrac{7}{6}\)
<=> x = \(\dfrac{1}{6}\)
@Nguyễn Hoàng Vũ
tinh hop li :
\(\frac{\left(\frac{1}{5}-\frac{2}{7}\right)\cdot\frac{3}{4}-\frac{3}{4}\cdot\left(\frac{1}{3}-\frac{2}{7}\right)}{\frac{1}{5}\cdot\frac{2}{7}-\frac{1}{3}\cdot\left(\frac{2}{7}+\frac{3}{9}\right)+\frac{3}{9}+\frac{1}{5}}\)
A = \(\left(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}\right)\)\(+\left(\frac{-6}{13}+\frac{1}{2}+1\frac{1}{3}\right)\)
B = \(0,75+\frac{2}{5}+\left(\frac{1}{9}-1\frac{2}{5}+\frac{5}{4}\right)\)
C =\(\left(-1\frac{1}{2}:\frac{3}{4}\right)\cdot\left(-4\frac{1}{2}\right)-\frac{1}{4}\)
D = \(1\frac{1}{4}\cdot\left(-0,7\cdot\frac{5}{4}-\frac{7}{8}\cdot\frac{14}{20}\right)\)
A= \(\left(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}\right)+\left(\frac{-6}{13}+\frac{1}{2}+\frac{4}{3}\right)\)
A= \(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}-\frac{6}{13}+\frac{1}{2}+\frac{4}{3}\)
A= \(\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{7}{13}+\frac{6}{13}\right)-\left(\frac{1}{3}-\frac{4}{3}\right)\)
A= \(1-1-\left(-1\right)\)
A= \(1\)
B= \(0,75+\frac{2}{5}+\left(\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\right)\)
B= \(\frac{3}{4}+\frac{2}{5}+\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\)
B= \(\left(\frac{3}{4}+\frac{5}{4}\right)+\left(\frac{2}{5}-\frac{7}{5}\right)+\frac{1}{9}\)
B= \(2-1+\frac{1}{9}\)
B= \(\frac{9}{9}+\frac{1}{9}\)
B= \(\frac{10}{9}\)
C= \(\left(\frac{-3}{2}.\frac{4}{3}\right).\left(\frac{-9}{2}\right)-\frac{1}{4}\)
C = \(-2.\left(\frac{-9}{2}\right)-\frac{1}{4}\)
C = \(9-\frac{1}{4}\)
C = \(\frac{36}{4}-\frac{1}{4}\)
C = \(\frac{35}{4}\)
D = \(\frac{5}{4}.\left(\frac{-7}{10}.\frac{5}{4}-\frac{7}{8}.\frac{7}{10}\right)\)
D = \(\frac{5}{4}.\left(\frac{-7}{8}-\frac{49}{80}\right)\)
D = \(\frac{-35}{32}-\frac{49}{64}\)
D = \(\frac{-70}{64}-\frac{49}{64}\)
D = \(\frac{-119}{64}\)
k mk nha ^_^
Câu A, B thì đúng rồi nhưng câu D, C thì mk chưa lm đc nhưng mk thây kết quả bài C, D hơi cao bởi bài này có đề là tính cách hợp lý nhất bn có cach làm nao khác giúp mk nha