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LV
.Câu 7,8,9,10 ạ!!! mjk đag gấp
1
20 tháng 8 2021
Xét tam giác BAM và tam giác CAM có:
MB=MC( gt)
AB=AC( gt)
AM chung
=> ΔBAM=ΔCAM( c.c.c)
=> \(\widehat{BAM}=\widehat{CAM}\) ( 2 góc tương ứng)
20 tháng 8 2021
MB=MC
AB=AC
AM
=> ΔBAM=ΔCAM( c.c.c)
=> ˆBAM=ˆCAMBAM^=CAM^
26 tháng 8 2021
Bài 2:
a: Xét ΔABD và ΔEBD có
BA=BE
\(\widehat{ABD}=\widehat{EBD}\)
BD chung
Do đó: ΔABD=ΔEBD
Suy ra: AD=ED
b: Ta có: ΔABD=ΔEBD
nên \(\widehat{BAD}=\widehat{BED}\)
mà \(\widehat{BAD}=90^0\)
nên \(\widehat{BED}=90^0\)
hay DE\(\perp\)BC
20 tháng 1
a: Xét ΔABD và ΔACD có
AB=AC
BD=CD
AD chung
Do đó: ΔABD=ΔACD
b: ΔABD=ΔACD
=>\(\widehat{ADB}=\widehat{ADC}\)
mà \(\widehat{ADB}+\widehat{ADC}=180^0\)(hai góc kề bù)
nên \(\widehat{ADB}=\widehat{ADC}=\dfrac{180^0}{2}=90^0\)
c: Ta có: \(\widehat{ADB}=90^0\)
=>AD\(\perp\)BC tại D
D là trung điểm của BC
=>\(DB=DC=\dfrac{BC}{2}=\dfrac{24}{2}=12\left(cm\right)\)
ΔADB vuông tại D
=>\(AD^2+DB^2=AB^2\)
=>\(AD^2=20^2-12^2=256\)
=>\(AD=\sqrt{256}=16\left(cm\right)\)
Xét ΔABC có
AD là đường trung tuyến
G là trọng tâm
Do đó: \(AG=\dfrac{2}{3}AD=\dfrac{2}{3}\cdot16=\dfrac{32}{3}\left(cm\right)\)
gấp gấp cứu e ạ
R
1
RY
24 tháng 4 2022
a. f(\(\dfrac{-1}{2}\)) = \(4.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)-2\)
= \(4.\dfrac{1}{4}-\left(\dfrac{-3}{2}\right)-\dfrac{4}{2}\)
= \(\dfrac{2}{2}+\dfrac{3}{2}-\dfrac{4}{2}\)
= \(\dfrac{1}{2}\)
19 tháng 7 2021
a, \(2^3.2^5=2^8=256\)
\(\left(-3\right)^9:\left(-3\right)^5=\left(-3\right)^4=81\)
\(\left(-6\right)^9.6^5=\left(-1\right)^9.6^9.6^5=\left(-1\right).6^{14}\\ \left(\dfrac{1}{2}\right)^5=\dfrac{1}{32}\)
b, \(\left(\dfrac{3}{5}\right)^6.\left(\dfrac{5}{3}\right)^6=\left(\dfrac{3}{5}. \dfrac{5}{3}\right)^6=1^6=1\\ \left(-\dfrac{7}{8}\right)^9:\left(\dfrac{7}{4}\right)^9=\left(-\dfrac{7}{8}:\dfrac{7}{4}\right)^9=\left(-\dfrac{1}{2}\right)^9=-\dfrac{1}{512}\\ \left(\left(-\dfrac{1}{2}\right)^2\right)^3=\left(\dfrac{1}{2}\right)^{...}\Rightarrow\left(\dfrac{1}{64}\right)=\left(\dfrac{1}{2}\right)...\Rightarrow\left(\dfrac{1}{64}\right)=\left(\dfrac{1}{2}\right)^6\)
c, \(\left(\dfrac{2}{3}\right)^8=\left(\left(\dfrac{2}{3}\right)^4\right)^{...}\Rightarrow\left(\left(\dfrac{2}{3}\right)^4\right)^2=\left(\left(\dfrac{2}{3}\right)^4\right)^{...}\Rightarrow\left(\dfrac{2}{3}\right)^8=\left(\left(\dfrac{2}{3}\right)^4\right)^2\\ \left(\dfrac{1}{3}\right)^{12}:\left(-\dfrac{3}{9}\right)^{12}=\left(\dfrac{1}{3}.\left(-3\right)\right)^{12}=\left(-1\right)^{12}=1\\ \left(\dfrac{1}{3}\right)^{12}:\left(\dfrac{1}{3}\right)^{10}=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)
\(A=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\cdot\left(5\cdot3-7\right)}=\dfrac{10-9}{15-7}=\dfrac{1}{8}\)
\(B=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2022\cdot2024}\right)\)
\(=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2023^2-1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2023^2}{\left(2023-1\right)\left(2023+1\right)}\)
\(=\dfrac{1}{2}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot2023}{1\cdot2\cdot3\cdot...\cdot2022}\cdot\dfrac{2\cdot3\cdot...\cdot2023}{3\cdot4\cdot...\cdot2024}\)
\(=\dfrac{1}{2}\cdot\dfrac{2023}{1}\cdot\dfrac{2}{2024}=\dfrac{1}{2024}\cdot\dfrac{2023}{1}=\dfrac{2023}{2024}\)
\(A=\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}\)
\(=\dfrac{5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^{10}.\left(2.3\right)^{19}-7.2^{29}.\left(3^3\right)^6}\)
\(=\dfrac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^{10}.2^{19}.3^{19}-7.2^{29}.3^{18}}\\ =\dfrac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{29}.3^{19}-7.2^{29}.3^{18}}\\ =\dfrac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{29}.3^{18}.\left(5.3-7\right)}\\ =\dfrac{10-9}{15-7}=\dfrac{1}{8}\)
\(---\)
\(B=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right)......\left(1+\dfrac{1}{2022.2024}\right)\\ =\dfrac{1}{2}.\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{2023^2}{2022.2024}\\ =\dfrac{1}{2}.\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{2023^2}{2022.2024}\\ =\dfrac{1}{2}.\dfrac{\left(2.3.4...2023\right)^2}{\left(1.2.3...2022\right)\left(3.4.5...2024\right)}\\ =\dfrac{1}{2}.\dfrac{2023.2}{2024}=\dfrac{2023}{2024}\)
#$\mathtt{Toru}$