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Ta có: \(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;\left|x+\frac{1}{12}\right|\ge0;...;\left|x+\frac{1}{110}\right|\ge0\)
=> VT \(\ge\)0
=>VP \(\ge\)0 => 11x \(\ge\)0 => x \(\ge\)0.
=> \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{6}\right|=x+\frac{1}{6};\left|x+\frac{1}{12}\right|=x+\frac{1}{12};...;\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\)
Phương trình <=> \(x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=11x\)
<=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\)
<=> \(10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)
<=> \(1-\frac{1}{11}=11x-10x\)
<=> \(\frac{10}{11}=x\)
<=> \(x=\frac{10}{11}\left(tm\right)\)
Bởi vì
\(\frac{1}{2}=\frac{1}{1.2};\frac{1}{6}=\frac{1}{2.3};...;\frac{1}{110}=\frac{1}{10.11}\)
nên từ \(\frac{1}{2}\)đến \(\frac{1}{110}\)chỉ có 10 số
nên chỉ có 10 x
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|=11x\)
Với mọi x ta có:
+) \(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\\.........\\\left|x+\frac{1}{110}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|\ge0\) \(\forall x.\)
Mà \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|=11x\)
\(\Rightarrow11x\ge0\)
\(\Rightarrow x\ge0.\)
Với \(x\ge0\) thì:
\(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|=x+\frac{1}{2}\\\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\\..........\\\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\end{matrix}\right.\)
\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{6}+...+x+\frac{1}{110}=11x\)
\(\Rightarrow11x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x-11x\)
\(\Rightarrow\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}=0x\) (vô lí).
\(\Rightarrow x\in\varnothing.\)
Vậy không tồn tại giá trị của x thỏa mãn yêu cầu đề bài.
Chúc bạn học tốt!
mình sửa đề chút nhé + \(\left|x+\frac{1}{110}\right|=11x\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
a) Dễ thấy VT > 0;mà VT=VP
=>VP > 0 => 4x > 0=> x > 0
=>\(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>BT đầu tương đương \(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{6}\right)=4x\)
\(=>3x+1=4x=>x=1\)
a) Để đẳng thức xảy ra thì: x>0 (vì: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|>0\) )
Khi đó: \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)
<=>x=1
Vậy x=1
b)Điều kiện: \(x\ne-3;-10;-21;-34\)
\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
=>x+34-x-3=x
<=>x=31 (nhận)
Vậy x=31
\(\frac{7}{8}.(\frac{2}{12}+\frac{4}{10})\)
\(\Rightarrow\frac{7}{8}.(\frac{10+24}{60})\)
\(\Rightarrow\frac{7}{8}.\frac{34}{60}=\frac{238}{480}\)
bt2
\(2.x-\frac{5}{4}=\frac{20}{15}\)
\(\Leftrightarrow2x=\frac{20}{15}+\frac{5}{4}\)
\(\Leftrightarrow2x=\frac{80+75}{60}\)
\(\Leftrightarrow2x=2,5\)
\(\Leftrightarrow x=1,25\)
.7/8.(1/6+2/5)=7/8.17/30=119/240
3/2-5/6:1/4+\(\sqrt{4}\)=3/2-10/3+2=1/6
2x=20/15+5/4
2x=31/12
x=31/12:2
x=31/24
ko bt nha thông cảm
\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)
\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)
\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
A) (x—1)2= | 1/4–1/2–3/4 |
(x—1)2= | 1/4–2/4–3/4 |
(x—1)2=|—1|
(x—1)2=1
==> (x—1)=1 hoặc (x—1)=-1
x=1+1 hoặc x—1=-1+1
x=2 hoặc x=0
b)(xx—8).(x2–15)<0
==> xx—8 <0 và x2> 0
Hay xx—8 >0 và x2<0
Mình chỉ biết tới đó thôi
\(a,\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)
\(\Rightarrow\left(x-1\right)^2=\left|-1\right|\)
\(\Rightarrow\left(x-1\right)^2=1\)
\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
vậy__
b, k bt