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B phai co ket qua chu khong co ket qua lam sao ma biet a = ?
thieu de bn oi,bai nay de that nhug thieu de thi sao chia dc
Ta có A = \(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
= \(\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)
= \(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
= \(\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\)
B = \(\dfrac{\dfrac{2}{29}-\dfrac{2}{39}+\dfrac{2}{49}}{\dfrac{23}{29}-\dfrac{23}{39}+\dfrac{23}{49}}=\dfrac{2\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}{23\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}=\dfrac{2}{23}\)
Lại có \(\dfrac{2}{23}>\dfrac{2}{24}=\dfrac{1}{12}\) hay A < B
Vậy A < B
a.
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow x\left(x+1\right).\left(x-1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
Đặt \(a=x^2+x-1\) , ta có pt:
\(\left(a+1\right)\left(a-1\right)-24=0\)
\(\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\)
\(\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)
*Với a = 5 ta được:
\(x^2+x-1=5\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
*Với a = -5 ta được:
\(x^2+x-1=-5\)
\(\Leftrightarrow x^2+x+4=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) ( loại)
Vậy pt có tập nghiệm là: \(s=\left\{-3;2\right\}\)
c)(ĐKXĐ: x khác 30;29)
\(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)
\(\Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{x-59}{30-x}+\dfrac{x-59}{29-x}\)
\(\Leftrightarrow x=59\)(tm) or \(\dfrac{1}{30}+\dfrac{1}{29}-\dfrac{1}{30-x}-\dfrac{1}{29-x}=0\)
\(\Leftrightarrow\dfrac{-x}{30\left(30-x\right)}+\dfrac{-x}{29\left(29-x\right)}=0\)
\(\Leftrightarrow x=0\)(tm) or \(\dfrac{1}{30\left(30-x\right)}+\dfrac{1}{29\left(29-x\right)}=0\)
\(\Leftrightarrow1741-59x=0\)
\(\Leftrightarrow x=\dfrac{1741}{59}\left(tm\right)\)
Vậy S={0;\(\dfrac{1741}{59}\);59}
Ta có :
\(A=3+3^2+3^3+........+3^{29}\)
\(\Rightarrow3A=3^2+3^3+...............+3^{29}+3^{30}\)
\(\Rightarrow3A-A=\left(3^2+3^3+........+3^{30}\right)-\left(3+3^3+................+3^{29}\right)\)
\(\Rightarrow2A=3^{30}-3\)
\(\Rightarrow A=\dfrac{3^{30}-3}{2}\)
Lại có :
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+................+\dfrac{1}{3^{29}}\)
\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+.............+\dfrac{1}{3^{28}}\)
\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+.......+\dfrac{1}{3^{28}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+..........+\dfrac{1}{3^{29}}\right)\)
\(\Rightarrow2B=1-\dfrac{1}{3^{29}}\)
\(\Rightarrow B=\dfrac{1-\dfrac{1}{3^{29}}}{2}\)
\(\dfrac{\Rightarrow A}{B}=\dfrac{\dfrac{3^{30}-3}{2}}{\dfrac{1-\dfrac{1}{3^{29}}}{2}}\)
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{29}}\)
\(3^{30}.B=3^{29}+3^{28}+...+3=A\)
\(\dfrac{A}{B}=\dfrac{3^{30}.B}{B}=3^{30}\)
a) \(-\frac{9}{13}.\frac{12}{29}-\frac{9}{13}.\frac{17}{29}+\frac{24}{13}\)
\(=-\frac{9}{13}\left(\frac{12}{29}+\frac{17}{29}\right)+\frac{24}{13}\)
\(=-\frac{9}{13}+\frac{24}{13}=\frac{15}{13}\)
b)\(\left(\frac{5}{3}-\frac{2}{11}\right):\frac{49}{33}+\left(-\frac{3}{7}\right)^2-\left|\frac{1}{7}-\frac{1}{14}\right|\)
\(=\left(\frac{5}{3}-\frac{2}{11}\right).\frac{33}{49}+\frac{9}{49}-\left|\frac{2}{14}-\frac{1}{14}\right|\)
\(=\frac{49}{33}.\frac{33}{49}+\frac{9}{49}-\frac{1}{14}\)
\(=1+\frac{9}{49}-\frac{1}{14}=\frac{49}{49}+\frac{9}{49}-\frac{1}{14}\)
\(=\frac{58}{49}-\frac{1}{14}=\frac{109}{98}\)
a) 19 + (29 - 9*37) - (63*9 - 29*99)
= 19 + 29 - 9*37 - 63*9 + 29*99
= 19 + 29(1 + 99) - 9(37 + 63)
= 19 + 29*100 - 9*100
= 19 + 100(29 - 9)
= 19 + 100*20
= 19 + 2000 = 2019
b) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)
= \(\frac{2^6+2^5+2^4+2^3+2^2+2+1}{2^7}\)
= \(\frac{64+32+16+8+4+2+1}{128}\) = \(\frac{127}{128}\)
a: Sửa đề: \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(A=2-\sqrt{3}+\sqrt{3}-1=1\)
c: \(C=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
d: Sửa đề: \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3\)
=6
a) \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(A=\left|4-\sqrt{15}\right|+\sqrt{15}\)
\(A=4-\sqrt{15}+\sqrt{15}\)
\(A=4\)
b) \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)}\)
\(B=\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\)
\(B=2-\sqrt{3}-1+\sqrt{3}\)
\(B=1\)
c) \(C=\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(C=\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{15}\cdot2+2^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot2+2^2}\)
\(C=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(C=\left|3\sqrt{5}-2\right|-\left|3\sqrt{5}+2\right|\)
\(C=3\sqrt{5}-2-3\sqrt{5}-2\)
\(C=-4\)
d) \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(D=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^3}\)
\(D=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(D=\left|2\sqrt{5}+3\right|-\left|2\sqrt{5}-3\right|\)
\(D=2\sqrt{5}+3-2\sqrt{5}+3\)
\(D=6\)
ta có a-1 phải là ước của -29 nên
\(a-1\in\left\{-29,-,1,1,29\right\}\Rightarrow a\in\left\{-28,0,2,30\right\}\)
tương ứng ta có cặp giá trị của a,b là
\(\left(-28,-1\right);\left(0,27\right);\left(2,-31\right);\left(30,-3\right)\)