giai phuong trinh 1/4(x+3)=3-1/2(x+1)-1/3(x+2)
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Câu hỏi của Phạm Thùy Dung - Toán lớp 7 - Học toán với OnlineMath
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a) \(\left(-5x+1\right)^2=\left(x-2\right)^2\)\(\Leftrightarrow\left(-5x+1\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[\left(-5x+1\right)-\left(x-2\right)\right]\left[\left(-5x+1\right)+\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(-5x+1-x+2\right)\left(-5x+1+x-2\right)=0\)
\(\Leftrightarrow\left(-6x+3\right)\left(-4x-1\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}-6x+3=0\\-4x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}-6x=-3\\-4x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-1}{4}\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{-1}{4};\frac{1}{2}\right\}\)
b) \(\left(-x-3\right)^2=\left(x+3\right)^2\)\(\Leftrightarrow\left(-x-3\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[\left(-x-3\right)-\left(x+3\right)\right]\left[\left(-x-3\right)+\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(-x-3-x-3\right)\left(-x-3+x+3\right)=0\)
\(\Leftrightarrow0=0\)
Vậy tập nghiệm của phương trình là \(S=ℝ\)
Ta thấy: 442=1936
452=2025
=>abcd=1936 (vì 1936<2020; 2025>2020)
=>1936+ef=2020
=>ef=84
Vậy a=1;b=9;c=3;d=6;e=8;f=4.
=2+2+...+2 (có 1006 số 2)
=2.1006
=2012
tk rồi mik tk lại
5x.5x+1.5x+2<100.................00:224
Có 24 số 0
53x.51.52<1024:2224
53x.53<524
53x<524:53
53x<521
=>3x=21
x=21:3
x=7\(\in\)N
Vậy x=7
Chúc bn học tốt
Bài 1 : Bài giải
\(-2\left(-3-4x\right)-3\left(3x-7\right)=31\)
\(6+8x-9x-21=31\)
\(-x-15=31\)
\(-x=31+15\)
\(-x=46\)
\(x=-46\)
b, \(10\left(x-7\right)=8\left(x-4\right)+x\)
\(10x-70=8x-32+x\)
\(10x-70=9x-32\)
\(10x-9x=-32+70\)
\(x=38\)
c, \(2\left|x-1\right|=3\cdot\left(5-1\right)\)
\(2\left|x-1\right|=15-3\)
\(2\left|x-1\right|=12\)
\(\left|x-1\right|=12\text{ : }2\)
\(\left|x-1\right|=6\)
\(\Rightarrow\orbr{\begin{cases}x-1=-6\\x-1=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-5\text{ ; }7\right\}\)
d, \(2\left(x+3\right)-2\left(x-3\right)=x\)
\(2\left(x+3-x-3\right)=x\)
\(2\cdot0=x\)
\(x=0\)
e, \(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-3\text{ ; }4\right\}\)
a, \(-2\left(-3-4x\right)-3\left(3x-7\right)=31\)
\(6+8x-9x-21=31\)
\(-x-15=31\)
\(-x=31+15\)
\(-x=46\)
\(x=-46\)
b, \(10\left(x-7\right)=8\left(x-4\right)+x\)
\(10x-70=8x-32+x\)
\(10x-70=9x-32\)
\(10x-9x=-32+70\)
\(x=38\)
c, \(2\left|x-1\right|=3\cdot\left(5-1\right)\)
\(2\left|x-1\right|=15-3\)
\(2\left|x-1\right|=12\)
\(\left|x-1\right|=12\text{ : }2\)
\(\left|x-1\right|=6\)
\(\Rightarrow\orbr{\begin{cases}x-1=-6\\x-1=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-5\text{ ; }7\right\}\)
d, \(2\left(x+3\right)-2\left(x-3\right)=x\)
\(2\left(x+3-x-3\right)=x\)
\(2\cdot0=x\)
\(x=0\)
e, \(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-3\text{ ; }4\right\}\)
\(A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^5}+...+\frac{2020}{5^{2020}}\)
\(\Rightarrow5A=1+\frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+...+\frac{2020}{5^{2019}}\)
\(\Rightarrow5A-A=4A=1+\left(\frac{2}{5}-\frac{1}{5}\right)+\left(\frac{3}{5^2}-\frac{2}{5^2}\right)+...+\left(\frac{2020}{5^{2019}}-\frac{2019}{5^{2019}}\right)-\frac{2020}{5^{2020}}\)
\(\Leftrightarrow4A=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}-\frac{2020}{5^{2020}}\)
\(B=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}\)
\(\Rightarrow5B=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}\)
\(\Rightarrow4B=1-\frac{1}{5^{2019}}\)
\(\Rightarrow B=\frac{1}{4}-\frac{1}{4.5^{2019}}\)
\(\Rightarrow4A=1+B-\frac{2020}{5^{2020}}\)
\(\Rightarrow A=\frac{5}{16}-\frac{1}{5^{2019}}\left(\frac{1}{4}+\frac{2020}{5}\right)=\frac{5}{16}-\frac{1617}{4.5^{2019}}\)
\(16>\frac{1617}{4.5^{2019}}\Rightarrow A=\frac{1}{4}+\left(\frac{1}{16}-\frac{1617}{4.5^{2019}}\right)>\frac{1}{4}\)
\(\frac{5}{16}< \frac{1}{3}\Rightarrow A< \frac{1}{3}\)
\(\Rightarrow\frac{1}{4}< A< \frac{1}{3}\left(Đpcm\right)\)
a. \(\left(x^2-2x+1\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow x^2-2x+1-3x^2+3x=0\)
\(\Leftrightarrow-2x^2+x+1=0\)
\(\Leftrightarrow-2x^2+2x-x+1=0\)
\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow-\left(2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)
Vậy \(x\in\left\{-\frac{1}{2};1\right\}\)
b. \(4\left(7x-3\right)-\left(7x^2-3x\right)=0\)
\(\Leftrightarrow4\left(7x-3\right)-x\left(7x-3\right)=0\)
\(\Leftrightarrow\left(4-x\right)\left(7x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\7x-3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{3}{7}\end{cases}}\)
Vậy \(x\in\left\{4;\frac{3}{7}\right\}\)
c.\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)
\(\Leftrightarrow\left(5-x\right)\left(2+3x\right)=\left(2-3x\right)\left(2+3x\right)\)
\(\Leftrightarrow\left(2+3x\right)\left(5-x-2+3x\right)=0\)
\(\Leftrightarrow\left(2+3x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2+3x=0\\2x+3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{3}{2}\end{cases}}\)
Vậy \(x\in\left\{-\frac{2}{3};-\frac{3}{2}\right\}\)
d. \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow7-4+4=-x+2x\)
\(\Leftrightarrow7=x\)
Vậy x = 7
e. \(\left(x-1\right)-\left(2x-1\right)=9\)
\(\Leftrightarrow x-1-2x+1=9\)
\(\Leftrightarrow-x=9\)
\(\Leftrightarrow x=-9\)
g. \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}}\)Mà : \(x^2+1\ge1>0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy x = -1