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Phần a vs phần b tính toán thông thường thôi mà bạn, vs 1 h/s lớp 7 thì ít nhất phải làm được chứ?? :((
a) \(x-\frac{4}{5}=\frac{7}{10}-\frac{3}{4}\)
\(\Leftrightarrow x-\frac{4}{5}=\frac{-1}{20}\)
\(\Leftrightarrow x=\frac{-1}{20}+\frac{4}{5}=\frac{15}{20}=\frac{3}{4}\)
b) \(2\frac{1}{3}-x=\frac{-5}{9}+2x\)
\(\Leftrightarrow2\frac{1}{3}-\frac{-5}{9}=2x+x\)
\(\Leftrightarrow3x=\frac{7}{3}+\frac{5}{9}\)
\(\Leftrightarrow3x=\frac{26}{9}\)
\(\Leftrightarrow x=\frac{26}{9}:3=\frac{26}{27}\)
d) .............................. ( Đề bài)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}\)\(-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Leftrightarrow-\frac{1}{x+3}=\frac{1}{2010}\)
\(\Leftrightarrow\frac{1}{-\left(x+3\right)}=\frac{1}{2010}\)\(\Leftrightarrow-\left(x+3\right)=2010\)
\(\Leftrightarrow-x-3=2010\) \(\Leftrightarrow-x=2010+3=2013\)
\(\Leftrightarrow x=-2013\)
Bạn tự kết luận nha!
c)
\(\frac{x+3}{2016}+\frac{x+2}{2017}=\frac{x+1}{2018}+\frac{x}{2019}\\ \Leftrightarrow\frac{x+3}{2016}+1+\frac{x+2}{2017}+1=\frac{x+1}{2018}+1+\frac{x}{2019}+1\\ \Leftrightarrow\frac{x+2019}{2016}+\frac{x+2019}{2017}-\frac{x+2019}{2018}-\frac{x+2019}{2019}=0\\ \Leftrightarrow\left(x+2019\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\\ \Rightarrow x-2019=0\\ \Rightarrow x=2019\)
b) để \(\left(x-7\right)^{x+2015}-\left(x-7\right)^{x+2016}=0\)
thì \(\left(x-7\right)^{x+2015}=\left(x-7\right)^{x+2016}\)
mà \(x+2015
nên \(x-7=x-7\Rightarrow x=7\)
bài a)
|2x+3|=x+2
2x+3=x+2 hoặc -(2x+3)=x+2
2x-x=2-3 -2x-3=x+2
1x=-1 -2x-x=2+3
x=-1 -3x =5
x=\(\frac{-5}{3}\)
Ta có \(\frac{2a+b+c}{b+c}=\frac{2b+c+a}{c+a}=\frac{2c+a+b}{a+b}\Rightarrow\frac{2a}{b+c}+1=\frac{2b}{a+c}+1=\frac{2c}{a+b}+1\)
=> \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{3}{2}\)
^_^
Bài 1: Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\)
\(\Rightarrow\hept{\begin{cases}a=2016k\\b=2017k\\c=2018k\end{cases}}\).Thay vào M,ta có:
\(M=4\left(2016k-2017k\right)\left(2017k-2018k\right)-\left(2018k-2016k\right)^2\)
\(=4.\left(-1k\right)\left(-1k\right)-\left(2k\right)^2\)
\(=4k^2-4k^2=0\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
a) \(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(c-a\right)^2+\left(b-c\right)^2=0\)
Ta có : \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(c-a\right)^2\ge0\\\left(b-c\right)^2\ge0\end{cases}}\)
\(\Rightarrow\left(a-b\right)^2+\left(c-a\right)^2+\left(b-c\right)^2=0\)
\(\Leftrightarrow a=b=c\)
a. \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ab-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\left(đpcm\right)\)
c) \(\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}+\frac{x+2}{2018}=\frac{x+2015}{5}+\frac{x+2016}{4}+\frac{x+2017}{3}+\frac{x+2018}{2}\)
Ta có VT + 4 = VP + 4
VT + 4 = \(\left(\frac{x+5}{2015}+1\right)+\left(\frac{x+4}{2016}+1\right)+\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+2}{2018}+1\right)\)
\(=\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}+\frac{x+2020}{2018}\)
\(=\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)\)
VP + 4 = \(\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)+\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}\right)\)
\(=\frac{x+2020}{5}+\frac{x+2020}{4}+\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(=\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)\)
Khi đó \(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)\)
=> \(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
Vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)
=> x + 2020 = 0
=> x = -2020
\(b,T=x\left(x-a\right)\left(x+a\right)\left(x+2a\right)+a^4\)
\(=x\left(x+a\right)\left(x-a\right)\left(x+2a\right)+a^4\)
\(=\left(x^2+ax\right)\left(x^2+ax-2a^2\right)+a^4\)
Đặt \(x^2+ax=t\)
\(\Rightarrow T=t\left(t-2a^2\right)+a^4\)
\(=t^2-2a^2t+a^4\)
\(=\left(t-a^2\right)^2\ge0\left(true\right)\)
Vậy \(T\ge0\)
c, Cộng mỗi hạng tử của 2 vế với 1
Đặt nhân tử chung là x + 2020
Từ đó tìm ra x = - 2020
a) a2 + b2 + c2 = ab + bc + ca
<=> 2( a2 + b2 + c2 ) = 2( ab + bc + ca )
<=> 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
<=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0
<=> ( a2 - 2ab + b2 ) + ( b2 - 2bc + c2 ) + ( c2 - 2ca + a2 ) = 0
<=> ( a - b )2 + ( b - c )2 + ( c - a )2 = 0
<=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow a=b=c\)( đpcm )
b) T = x( x - a )( x + a )( x + 2a ) + a4
= [ x( x + a ) ][ ( x - a )( x + 2a ) ] + a4
= [ x2 + ax ][ x2 + ax - 2a2 ] + a4
Đặt t = x2 + ax
=> T = t( t - 2a2 ) + a4
= t2 - 2a2t + a4
= ( t - a2 )2
= ( x2 + ax - a2 )2 \(\ge0\forall x,a\in R\)( đpcm )
c) \(\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}+\frac{x+2}{2018}=\frac{x+2015}{5}+\frac{x+2016}{4}+\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+5}{2015}+1\right)+\left(\frac{x+4}{2016}+1\right)+\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+2}{2018}+1\right)=\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)+\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}+\frac{x+2020}{2018}=\frac{x+2020}{5}+\frac{x+2020}{4}+\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(\Leftrightarrow\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}+\frac{x+2020}{2018}-\frac{x+2020}{5}-\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
Vì \(\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)\ne0\)
=> x + 2020 = 0 => x = -2020
P/s: Lười check lại ... Sai sót gì mong bạn bỏ qua :)