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Đầu tiên , ta cộng các phần nguyên lại với nhau trước :
( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) + ( \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{8}{72}+\frac{1}{90}+\frac{1}{10}\)
= 45 + \(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{42}+\frac{1}{72}\right)+\left(\frac{1}{10}+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{90}\right)+\frac{1}{56}\)
= 45 +
tới đây tớ chịu , các cậu giúp với
Đầu tiên , cộng các phần nguyên lại với nhau , ta có :
( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) + ( \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\))
= 45 + \(\left(\frac{1}{6}+\frac{1}{30}\right)+\frac{1}{2}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi cộng trong ngoặc , ta được 6 / 30 , rút gọn tối giản còn 1 / 5
= 45 + \(\left(\frac{1}{5}+\frac{1}{20}\right)+\frac{1}{2}+\frac{1}{12}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi cộng trong ngoặc và rút gọn tối giản , ta được 1 / 4
= 45 + \(\left(\frac{1}{4}+\frac{1}{2}\right)+\frac{1}{12}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi cộng trong ngoặc rồi rút gọn , ta được 3 / 4
= 45 + \(\left(\frac{3}{4}+\frac{1}{12}\right)+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
rút gọn lại ta được 5 / 6
= 45 + \(\left(\frac{5}{6}+\frac{1}{42}\right)+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
rút gọn tối giản ra 6 / 7
= 45 + \(\left(\frac{6}{7}+\frac{1}{56}\right)+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi tính trong ngoặc rút gọn được 7 / 8
= 45 + \(\left(\frac{7}{8}+\frac{1}{72}\right)+\frac{1}{90}+\frac{1}{10}\)
tính trong ngoặc rồi rút gọn ra 8 / 9
= 45 + \(\left(\frac{8}{9}+\frac{1}{90}\right)+\frac{1}{10}\)
cũng rút gọn tiếp ta được 9 / 10
= 45 + \(\left(\frac{9}{10}+\frac{1}{10}\right)\)
= 45 + 1
= 46
1: =>\(5^{2x-3}=5^2\cdot3+5^2\cdot2=5^2\cdot5=5^3\)
=>2x-3=3
=>2x=6
=>x=3
2: \(41-2^{x+1}=9\)
=>\(2^{x+1}=32\)
=>x+1=5
=>x=4
3: =>\(4^{x+2}=65-1=64\)
=>x+2=3
=>x=1
\(5^{2x-3}-2.5^2=5^2.3\\ 5^{2x-3}=5^2.3+5^2.2\\ 5^{2x-3}=5^2.\left(3+2\right)\\ 5^{2x-3}=5^2.5\\ 5^{2x-3}=5^3\\ \Rightarrow2x-3=3\\ 2x=3+3\\ 2x=6\\ x=\dfrac{6}{2}\\ Vậy:x=3\)
a/ \(\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}\)
\(=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+4}\)
Vậy..
b/ \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+5}\)
Vậy..
x2-4x+7 = 0 ⇔ x2 -4x + 4 + 3 = 0
⇔ (x-2)2+3=0 ⇔ (x-2)2=-3 (vô lí)
Vậy pt vô nghiệm
*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm
Ta có: \(x^2-4x+7=0\)
\(\Leftrightarrow x^2-4x+4+3=0\)
\(\Leftrightarrow\left(x-2\right)^2+3=0\)
mà \(\left(x-2\right)^2+3\ge3>0\forall x\)
nên \(x\in\varnothing\)(đpcm)
ĐK : \(x\ne-2.-3;-4;-5;-6\)
\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\Leftrightarrow\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\Leftrightarrow x^2+8x-20=0\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\Leftrightarrow x=2;x=-10\)( tmđkxđ )
Vậy tập nghiệm phương trình là S = { -10 ; 2 }
ĐKXĐ \(x\notin\left\{-2;-3;...;-6\right\}\)
Phương trình tương đương với:
\(\dfrac{1}{\left(x^2+2x\right)+\left(3x+6\right)}+\dfrac{1}{\left(x^2+3x\right)+\left(4x+12\right)}+\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{\left(x+3\right)-\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}+\dfrac{\left(x+4\right)-\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}+\dfrac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\dfrac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{4}{32}\\ \Rightarrow\left(x+2\right)\left(x+6\right)=32\\\Leftrightarrow x^2+8x-20=0\\ \Leftrightarrow\left(x+10\right)\left(x-2\right)=0\\ \Leftrightarrow\begin{matrix}x=2\\x=-10\end{matrix}\left(t.m\right)\)
\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+3\right)}+\frac{1}{\left(x+3\right)}-...-\frac{1}{x+6}+\frac{1}{\left(x+6\right)}-\frac{1}{\left(x+7\right)}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+7}=\frac{1}{2}\Leftrightarrow\frac{6}{\left(x+1\right)\left(x+7\right)}=\frac{1}{2}\)\(\Leftrightarrow x^2+8x+7=12\Leftrightarrow\left(x+4\right)^2-21=0\Leftrightarrow\left(x+4-\sqrt{21}\right)\left(x+4+\sqrt{21}\right)=0\Rightarrow\left[{}\begin{matrix}x=-4+\sqrt{21}\\x=-4-\sqrt{21}\end{matrix}\right.\)
\(\left(2x-1\right)^3-4x^2\left(2x-3\right)=5.\)
\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)
\(\Leftrightarrow6x-1=5\)
\(\Leftrightarrow6x=6\)
\(\Leftrightarrow x=1\)
\(\left(x+4\right)^3-x^2\left(x+12\right)=15.\)
\(\Leftrightarrow x^3+12x^2+48x+64-x^3-12x^2=15\)
\(\Leftrightarrow48x+64=15\)
\(\Leftrightarrow48x=-49\)
\(\Leftrightarrow x=\frac{-49}{48}\)
