tính : 19/21 : 273/228 =
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Giải Phương Trình Sau:
\(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
PT $\Leftrightarrow \frac{x-342}{15}-1+\frac{x-323}{17}-2+\frac{x-300}{19}-3+\frac{x-273}{21}-4=0$
$\Leftrightarrow \frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0$
$(x-357)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0$
Dễ thấy: $\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\neq 0$
$\Rightarrow x-357=0$
$\Rightarrow x=357$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
<=>\(\dfrac{x-342}{15}-1+\dfrac{x-323}{17}-2+\dfrac{x-300}{19}-3+\dfrac{x-273}{21}-4=0\)
<=>\(\dfrac{x-357}{15}+\dfrac{x-357}{17}+\dfrac{x-357}{19}+\dfrac{x-357}{21}=0\)
<=>\(\left(x-357\right)\left(\dfrac{1}{15}+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}\right)=0\)
vì 1/15+1/17+1/19+1/21 khác 0=>x-357=0<=>x=357
vậy..................
chúc bạn học tốt ^^
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(34 - 53 + 28) - (34 - 53 - 21)
= 34 - 53 + 28 - 34 + 53 + 21
= (34 - 34) - (53 - 53) + (28 + 21)
= 0 - 0 + 49
= 49
(23 - 15 + 38) - (-47 - 15 + 23)
= 23 - 15 + 38 + 47 + 15 - 23
= (23 - 23) - (15 - 15) + (38 + 47)
= 85
(125 + 49 - 37) - (49 + 125 - 228)
= 125 + 49 - 37 - 49 - 125 + 228
= (125 - 125) + (49 - 49) - (37 - 228)
= -(-191) = 191
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề; \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
\(\Leftrightarrow x-357=0\)
hay x=357
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^4+3x^3+8x^2+6x+5=0\)
\(\Leftrightarrow2x^4+2x^3+2x^2+x^3+x^2+x+5x^2+5x+5=0\)
\(\Leftrightarrow2x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+5\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(2x^2+x+5\right)=0\)
Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(2x^2+x+5=2\left[\left(x+\frac{1}{4}\right)^2+\frac{39}{16}\right]>0\forall x\)
Vậy tập nghiệm của pt là \(S=\varnothing\)
b, \(\frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}=10\)
\(\Leftrightarrow\left(\frac{x-342}{15}-1\right)+\left(\frac{x-323}{17}-2\right)+\left(\frac{x-300}{19}-3\right)+\left(\frac{x-273}{21}-4\right)=0\)
\(\Leftrightarrow\frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0\)
\(\Leftrightarrow\left(x-357\right)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0\)
\(\Leftrightarrow x-357=0\Leftrightarrow x=357\)
Vậy tập nghiệm của pt: \(S=\left\{357\right\}\)
\(\frac{x-342}{15}\) + \(\frac{x-323}{17}\) + \(\frac{x-300}{19}\) + \(\frac{x-273}{21}\) =10
giải pt
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}=10\)
=> \(\left(\frac{x-342}{15}-1\right)+\left(\frac{x-323}{17}-2\right)+\left(\frac{x-300}{19}-3\right)+\left(\frac{x-273}{21}-4\right)=0\)
=> \(\frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0\)
=> \(\left(x-357\right)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0\)
Vì \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\ne0\)
=> x - 357 = 0
=> x = 357
Vậy x = 357
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(19^{21}+19^{22}+19^{23}\right):\left(19^{20}+19^{21}+19^{22}\right)\)
\(=19^{21}.\left(1+19+19^2\right):19^{20}:\left(1+19+19^2\right)=19\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left|157\dfrac{13}{27}-273\dfrac{7}{19}\right|-96\dfrac{14}{27}+15\dfrac{12}{19}\)
\(=273\dfrac{7}{19}-153\dfrac{13}{27}-96\dfrac{14}{27}+15\dfrac{12}{19}\)
\(=\left(273+15+\dfrac{7}{19}+\dfrac{12}{19}\right)-\left(153+96+\dfrac{13}{27}+\dfrac{14}{27}\right)\)
\(=289-250=39\)
TL
19/21:273/228=1444/1911
nha
HT