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a: góc yAt'=180 độ-60 độ=120 độ
góc yAt'=góc yOx
mà hai góc này đồng vị
nên At'//Ox
b: góc mOA=góc xOy/2=60 độ
góc nAO=góc OAt/2=60 độ
=>góc mOA=góc nAO
=>Om//An
Thì bạn phân tích ra thừa số nguyên tố á
Bấm số đó vào máy
Rồi bấm ShiFT+FACT nha
Rồi sau đó thấy tự động phân tích số mũ cho mình luôn à
\(A=\sqrt[]{50}+\sqrt[]{65}\Rightarrow A^2=50+65+2\sqrt[]{50.65}=115+2\sqrt[]{5.10.5.13=}115+10\sqrt[]{130}\left(1\right)\)
\(B=\sqrt[]{15}+\sqrt[]{115}\Rightarrow B^2=15+115+2\sqrt[]{15.115}=15+115+2\sqrt[]{3.5.5.23}=15+115+10\sqrt[]{69}\left(2\right)\)Ta có \(10\sqrt[]{130}< 10\sqrt[]{69.2}=10\sqrt[]{2}\sqrt[]{69}< 15+10\sqrt[]{69}\left(3\right)\)
\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow A^2< B^2\Rightarrow A< B\)
\(\Rightarrow\sqrt[]{50}+\sqrt[]{65}< \sqrt[]{15}+\sqrt[]{115}\)
So sánh gì thế em, em nhập đủ đề vào hi
`@` `\text {Ans}`
`\downarrow`
`7)`
`2 \sqrt {1 - 2x} - 5 = 0`
`=> 2 \sqrt {1 - 2x} = 5`
`=> \sqrt {1 - 2x} = 5/2`
`=> 1 - 2x = (5/2)^2`
`=> 2x = 1 - (5/2)^2`
`=> 2x = 1 - 25/4`
`=> 2x = -21/4`
`=> x = -21/4 \div 2`
`=> x =-21/8`
Vậy, `x = -21/8`
`8)`
`(x^2 + 1)*(2x^2 - 3) = 0`
`=>`
`\text {TH1: } x^2 + 1 = 0`
`=> x^2 = 0 - 1`
`=> x^2 = -1 (\text {vô lý})`
Vậy, x không có gtr thỏa mãn
`\text {TH2: } 2x^2 - 3 = 0`
`=> 2x^2 = 0 + 3`
`=> 2x^2 = 3`
`=> x^2 = 3/2`
`=> x = +- \sqrt {3/2}`
Vậy, `x \in`\(\left\{-\sqrt{\dfrac{3}{2}};\sqrt{\dfrac{3}{2}}\right\}\)
2: =>2*căn 1-2x=5
=>căn 1-2x=5/2
=>1-2x=25/4
=>2x=1-25/4=-21/4
=>x=-21/8
8: =>2x^2-3=0 hoặc x^2+1=0(vô lý)
=>2x^2=3
=>x^2=3/2
=>\(x=\pm\dfrac{\sqrt{6}}{2}\)
9: =>|3/5*căn x-1/20|=3/4+1/5=19/20
=>3/5*căn x-1/20=-19/20 hoặc 3/5*căn x-1/20=19/20
=>3/5*căn x=-18/20(vô lý) hoặc 3/5*căn x=1
=>căn x=5/3
=>x=25/9
N = -3\(x\)(4\(x^2\) +5) - 2\(x^2\).(4 -6\(x\)) + 9\(x^2\)
Vì |\(x\)| = 1; ⇔ (|\(x\)|)2 = \(x^2\) = 1
Thay \(x^2\) = 1 vào N ta có:
N = -3\(x\)(4\(x^2\) + 5) - 2\(x^2\).(4 -6\(x\)) + 9\(x^2\)
N = -3\(x\)( 4 + 5) - 2(4 - 6\(x\)) + 9
N = -3\(x\).9 - 8 + 12\(x\) + 9
N = - 27\(x\) + 12\(x\) + 1
N = -15\(x\) + 1
|\(x\)| =1 ⇒ \(x\) = 1; -1
thay \(x\) = 1 vào N = -15\(x\) + 1 = -15 + 1 = - 14
Thay \(x\) = -1 vào N = -15\(x\) + 1 = (-15).(-1) + 1 = 16
a) \(3^{54}\)
\(2^{200}=4^{100}>3^{54}\)
\(\Rightarrow3^{54}< 2^{200}\)
b) \(15^{12}=3^{12}.5^{12}\)
\(1^3.125^3=\left(5^3\right)^3=5^9< 3^{12}.5^{12}\)
\(\Rightarrow15^{12}>1^3.125^3\)
c) \(78^{12}-78^{11}=78^{11}.\left(7-1\right)=78^{11}.6\)
\(78^{11}-78^{10}=78^{10}.\left(7-6\right)=78^{10}.6< 78^{11}.6\)
\(\Rightarrow78^{12}-78^{11}>78^{11}-78^{10}\)
d) \(72^{45}-72^{44}=72^{44}.\left(72-1\right)=72^{44}.72>27^{44}\)
\(\Rightarrow72^{45}-72^{44}>27^{44}\)
e) \(3^{39}=\left(3^3\right)^{13}=27^{13}>11^{11}\)
\(\Rightarrow3^{39}>11^{11}\)
\(\left|3x-2\right|=0\\ \Leftrightarrow3x-2=0\\ \Leftrightarrow3x=2\\ \Leftrightarrow x=\dfrac{2}{3}\\ ---\\ \left|3x-2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=4\\-\left(3x-2\right)=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=6\\-3x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\sqrt{3x+1}=\dfrac{2}{3}\left(ĐKXĐ:3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}\right)\\ \Leftrightarrow\left|3x+1\right|=\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}\\ \Leftrightarrow\left[{}\begin{matrix}3x+1=\dfrac{4}{9}\\-3x-1=\dfrac{4}{9}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{5}{9}\\-3x=\dfrac{13}{9}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{9}:3=-\dfrac{5}{27}\left(nhận\right)\\x=\dfrac{13}{9}:\left(-3\right)=\dfrac{-13}{27}\left(loại\right)\end{matrix}\right.\\ Vậy:x=-\dfrac{5}{27}\)
a) 0,25=(0,5)2; \(\dfrac{8}{125}=\left(\dfrac{2}{5}\right)^3\)
\(-\dfrac{27}{125}=\left(-\dfrac{3}{5}\right)^3\); \(\dfrac{121}{169}=\left(\dfrac{11}{13}\right)^2\)
b) \(\left(\dfrac{1}{4}\right)^3=\left[\left(\dfrac{1}{2}\right)^2\right]^3=\left(\dfrac{1}{2}\right)^6\);
\(\left(\dfrac{1}{8}\right)^5=\left[\left(\dfrac{1}{2}\right)^3\right]^5=\left(\dfrac{1}{2}\right)^{15}\)
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\(a\)) \(\text{0,25 = }\left(\dfrac{1}{2}\right)^2\) \(;\) \(\dfrac{\text{8}}{\text{125}}=\left(\dfrac{2}{5}\right)^3\)\(;\) \(\dfrac{\text{-27}}{\text{1}\text{12}}=\left(\dfrac{-3}{5}\right)^3\) \(;\) \(\dfrac{\text{121}}{\text{169}}=\left(\dfrac{11}{13}\right)^2\)
\(b\)) \(\text{1/2 = (1/2)^1}\) \(;\) \(\text{(1/4)^3 = 1/64 = (1/4)^6}\) \(;\) \(\text{(1/8)^5 = 1/32768 = (1/8)^15}\)