Tính giá trị biểu thức
42,6 + 1,38 x 0,7 - 2,04
Làm chi tiết ra cho mình luôn nka, cảm ơn ạ!:)^^
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NM
Nguyễn Minh Quang
Giáo viên
16 tháng 12 2020
Ta có hai trường hợp như sau :
TH1
\(x-2016\ge0\Leftrightarrow x\ge2016\) thì \(A=x-2016+x-1=2x-2017\ge2.2016-2017=2015\)
TH2
\(x-2016\le0\Leftrightarrow x\le2016\) thì \(A=2016-x+x-1=2015\)
vì vậy GTNN của A=2015
dấu bằng xảy ra khi \(x\le2016\)
NT
3
NC
2
9 tháng 11 2021
1, Áp dụng PTG: \(AC=\sqrt{BC^2-AB^2}=8\left(cm\right)\)
Áp dụng HTL: \(\left\{{}\begin{matrix}CH=\dfrac{AC^2}{BC}=6,4\left(cm\right)\\AH=\dfrac{AB\cdot AC}{BC}=4,8\left(cm\right)\end{matrix}\right.\)
\(\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{4}{5}\approx\sin53^0\\ \Rightarrow\widehat{B}\approx53^0\\ \Rightarrow\widehat{C}\approx90^0-53^0=37^0\)
2,
a, Áp dụng HTL: \(\left\{{}\begin{matrix}AD\cdot AB=AH^2\\AE\cdot AC=AH^2\end{matrix}\right.\Rightarrow AD\cdot AB=AE\cdot AC\)
b, \(AD\cdot AB=AE\cdot AC\Rightarrow\dfrac{AD}{AC}=\dfrac{AE}{AB}\Rightarrow\Delta ABC\sim\Delta AED\left(c.g.c\right)\)
2 tháng 10 2020
\(\frac{45^{10}.5^{20}}{75^{15}}\)
\(=\frac{\left(15.3\right)^{10}.5^{20}}{\left(15.5\right)^{15}}\)
\(=\frac{15^{10}.3^{10}.5^{20}}{15^{15}.5^{15}}\)
\(=\frac{3^{10}.5^5}{15^5}=\frac{3^{10}.5^5}{3^5.5^5}=3^5=243\)
2 tháng 10 2020
\(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(9.5\right)^{10}.5^{20}}{\left(3.5.5\right)^{15}}=\frac{9^{10}.5^{10}.5^{20}}{3^{15}.5^{15}.5^{15}}=\frac{9^{10}.5^{30}}{3^{15}.5^{30}}=\frac{9^{10}}{3^{15}}=243\)
9P
Cho mình xin lời giải chi tiết luôn ạ cực gấp xin cảm ơn ạ
1
10 tháng 1 2022
\(=\dfrac{3x-6+5x+10+3x-26}{\left(x-2\right)\left(x+2\right)}=\dfrac{11x-22}{\left(x-2\right)\left(x+2\right)}=\dfrac{11}{x+2}\)
NC
2
9 tháng 11 2021
Bài 1:
\(a,A=6\sqrt{2}-6\sqrt{2}+2\sqrt{5}=2\sqrt{5}\\ b,B=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{3}+\sqrt{2}\\ c,=2\sqrt{3}-6\sqrt{3}+15\sqrt{3}-4\sqrt{3}=7\sqrt{3}\\ d,=1+6\sqrt{3}-\sqrt{3}-1=5\sqrt{3}\\ e,=4\sqrt{2}+\sqrt{2}-6\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
Bài 2:
\(a,ĐK:x\ge\dfrac{3}{2}\\ PT\Leftrightarrow\sqrt{2x-3}=5\Leftrightarrow2x-3=25\Leftrightarrow x=14\\ b,PT\Leftrightarrow x^2=\sqrt{\dfrac{98}{2}}=\sqrt{49}=7\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\end{matrix}\right.\\ c,ĐK:x\ge3\\ PT\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+1\right)=0\\ \Leftrightarrow\sqrt{x-3}=0\left(\sqrt{x+3}+1>0\right)\\ \Leftrightarrow x=3\\ d,ĐK:x\ge1\\ PT\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\\ \Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\left(tm\right)\\ e,PT\Leftrightarrow2x-1=16\Leftrightarrow x=\dfrac{17}{2}\\ f,PT\Leftrightarrow\left|2x-1\right|=\sqrt{3}-1\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{3}-1\\2x-1=1-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}\\x=\dfrac{2-\sqrt{3}}{2}\end{matrix}\right.\)
9 tháng 11 2021
Bài 3:
\(a,Q=\dfrac{1+5}{3-1}=3\\ b,P=\dfrac{x+\sqrt{x}-6+x-2\sqrt{x}-3-x+4\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ c,M=\dfrac{\sqrt{x}}{\sqrt{x}-3}\cdot\dfrac{3-\sqrt{x}}{\sqrt{x}+5}=\dfrac{-\sqrt{x}}{\sqrt{x}+5}\)
Vì \(-\sqrt{x}\le0;\sqrt{x}+5>0\) nên \(M< 0\)
Do đó \(\left|M\right|>\dfrac{1}{2}\Leftrightarrow M< -\dfrac{1}{2}\Leftrightarrow-\dfrac{\sqrt{x}}{\sqrt{x}+5}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-\sqrt{x}-5}{2\left(\sqrt{x}+5\right)}< 0\Leftrightarrow\sqrt{x}-5< 0\left(\sqrt{x}+5>0\right)\\ \Leftrightarrow0\le x< 25\)
Bài 4:
\(a,A=\dfrac{16+2\cdot4+5}{4-3}=29\\ b,B=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ c,P=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}+1}\\ P=\dfrac{\left(\sqrt{x}+1\right)^2+4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}\\ P\ge2\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}=2\sqrt{4}=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\Leftrightarrow\sqrt{x}+1=2\Leftrightarrow x=1\left(tm\right)\)
BG
3
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2021
\(=\sqrt{7-2\sqrt{21}+3}+\sqrt{7+2\sqrt{21}+3}\)
\(=\sqrt{\sqrt{7}^2-2\sqrt{7}.\sqrt{3}+\sqrt{3}^2}+\sqrt{\sqrt{7}^2+2\sqrt{7}.\sqrt{3}+\sqrt{3}^2}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)
\(=2\sqrt{7}\)
26 tháng 7 2021
\(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)
\(=2\sqrt{7}\)
26 tháng 7 2021
b) Ta có: \(B=\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)
\(=2\sqrt{7}\)
d) Ta có: \(D=\sqrt{x^2-6x+9}-x\)
\(=\left|x-3\right|-x\)
\(=\left[{}\begin{matrix}x-3-x=-3\left(x\ge3\right)\\3-x-x=-2x+3\left(x< 3\right)\end{matrix}\right.\)
\(42,6+1,38\times0,7-2,04.\)
\(=42,6+0,966-2,04\)
\(=43,566-2,04\)
\(=41,526\)
TL
42,6 + 1,38 x 0,7 - 2,04
=42,6 + 0,966 - 2,04
=41,526
HT Ạ
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