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LN
1
DD
Đoàn Đức Hà
Giáo viên
16 tháng 12 2021
\(S=\frac{1}{x\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\)
\(=\frac{1}{2}\left[\frac{2}{x\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{2}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\right]\)
\(=\frac{1}{2}\left[\frac{x+2-x}{x\left(x+1\right)\left(x+2\right)}+\frac{\left(x+3\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{\left(x+10\right)-\left(x+8\right)}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\right]\)
22 tháng 7 2021
trung bình cộng của các số 545,328,624,295 là bao nhiêu
26 tháng 2 2020
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\left(x\ne1\right)\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x}{x^2+x+1}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\left(x^2+x+1-3x^2-2x^2+2x\right)=0\)
\(\Leftrightarrow-4x^2+3x+1=0\left(\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\ne0\right)\)
\(\Leftrightarrow-4x^2+4x-x+1=0\)
\(\Leftrightarrow-4x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-4x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\-4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\-4x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\left(loại\right)\\x=\frac{-1}{4}\end{cases}}}\)
Vậy \(x=\frac{-1}{4}\)
22 tháng 7 2018
\(a,\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right)\)
\(=\left(x-2\right)^3-x\left(x^2-1\right)+6x^2-18x\)
\(=x^3-6x^2+12x-8-x^3+x+6x^2-18x\)
\(=-5x\)
Các câu còn lại lm tương tự nhé
9 tháng 1 2021
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
PN
16 tháng 5 2021
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
17 tháng 6 2022
a: \(\dfrac{x-1}{x^2-x+1}-\dfrac{x+1}{x^2+x+1}=\dfrac{10}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)-x\left(x+1\right)\left(x^2-x+1\right)=10\)
\(\Leftrightarrow x\left(x^3-1\right)-x\left(x^3+1\right)=10\)
=>-2x=10
hay x=-5
d: \(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+7\right)\left(x+8\right)}=\dfrac{1}{14}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+8}=\dfrac{1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(x+8\right)=14\left(x+8\right)-14\left(x+1\right)\)
\(\Leftrightarrow x^2+9x+8=14x+112-14x-14=98\)
\(\Leftrightarrow x^2+9x-90=0\)
\(\Leftrightarrow x\in\left\{6;-15\right\}\)
AH
Akai Haruma
Giáo viên
21 tháng 7 2021
Lời giải:
$(x-10)^2-2(x-10)(5x-3)+(5x-3)^2$
$=[(x-10)-(5x-3)]^2=(-4x-7)^2=(4x+7)^2$
$(x-2)^3-x(x-3)^2=x^3-6x^2+12x-8-x(x^2-6x+9)$
$=x^3-6x^2+12x-8-x^3+6x^2-9x$
$=3x-8$
$(x-2)(x^2+2x+4)-x(x+1)(x-1)$
$=(x^3-2^3)-x(x^2-1)$
$=x^3-8-x^3+x=x-8$
$(x+3)(x^2-3x+9)-x(x^2-1)=(x^3+3^3)-x^3+x$
$=x+27$
\(\dfrac{1}{x+1}:\dfrac{x+2}{x+1}:\dfrac{x+3}{x+2}:...:\dfrac{x+10}{x+9}\)
\(=\dfrac{1}{x+1}\cdot\dfrac{x+1}{x+2}\cdot\dfrac{x+2}{x+3}\cdot...\cdot\dfrac{x+9}{x+10}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\cdot...\cdot\left(x+9\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\cdot...\cdot\left(x+9\right)\left(x+10\right)}\)
\(=\dfrac{1}{x+10}\)