câu 2a) \(\dfrac{1}{11}\left(1-3x\right)=\dfrac{1}{5}x-\dfrac{1}{2}\)
\(\dfrac{1}{11}-\dfrac{3}{11}x=\dfrac{1}{5}x-\dfrac{1}{2}\\ \Leftrightarrow-\dfrac{1}{5}x-\dfrac{3}{11}x=-\dfrac{1}{11}-\dfrac{1}{2}\\ -\dfrac{26}{55}x=-\dfrac{13}{22}\Rightarrow x=\dfrac{5}{4}\)
b) \(\sqrt{\dfrac{9}{4}}-\left|2x+1\right|=0,75\)
\(\dfrac{3}{2}-\left(2x+1\right)=0,75\\ \dfrac{3}{2}-2x-1=0,75\\ -2x=\dfrac{1}{4}\\ x=-\dfrac{1}{8}\)
Câu 1:
1: \(\dfrac{3}{7}:\left(-\dfrac{4}{5}+\dfrac{1}{2}\right)-\dfrac{3}{7}:\dfrac{-4}{5}\)
\(=\dfrac{3}{7}:\left(-\dfrac{8}{10}+\dfrac{5}{10}\right)+\dfrac{3}{7}\cdot\dfrac{5}{4}\)
\(=\dfrac{3}{7}\cdot\dfrac{-10}{3}+\dfrac{3}{7}\cdot\dfrac{5}{4}=\dfrac{3}{7}\left(-\dfrac{10}{3}+\dfrac{5}{4}\right)\)
\(=\dfrac{3}{7}\left(-\dfrac{40}{12}+\dfrac{15}{12}\right)=\dfrac{3}{7}\cdot\dfrac{-25}{12}=\dfrac{-25}{28}\)
2: \(\left(-5\right)^{2021}\cdot x^{2022}-\left|y+\dfrac{3}{4}\right|>=0\)
=>\(-5^{2021}\cdot x^{2022}-\left|y+\dfrac{3}{4}\right|>=0\)
=>\(5^{2021}\cdot x^{2022}+\left|y+\dfrac{3}{4}\right|< =0\)
mà \(5^{2021}\cdot x^{2022}+\left|y+\dfrac{3}{4}\right|>=0\forall x,y\)
nên \(\left\{{}\begin{matrix}x=0\\y+\dfrac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-\dfrac{3}{4}\end{matrix}\right.\)
\(P=\left(-\dfrac{21}{22}\right)^x-\sqrt{y+1}\)
\(=\left(-\dfrac{21}{22}\right)^0-\sqrt{-\dfrac{3}{4}+1}=1-\sqrt{\dfrac{1}{4}}=1-\dfrac{1}{2}=\dfrac{1}{2}\)
3: \(2a-b=\dfrac{2}{3}\left(a+b\right)\)
=>\(2a-b=\dfrac{2}{3}a+\dfrac{2}{3}b\)
=>\(2a-\dfrac{2}{3}a=\dfrac{2}{3}b+b\)
=>\(\dfrac{4}{3}a=\dfrac{5}{3}b\)
=>4a=5b
=>\(\dfrac{a}{5}=\dfrac{b}{4}=k\)
=>a=5k; b=4k
\(M=\dfrac{a^4+5^4}{b^4+4^4}=\dfrac{\left(5k\right)^4+5^4}{\left(4k\right)^4+4^4}=\dfrac{5^4\left(k^4+1\right)}{4^4\left(k^4+1\right)}=\left(\dfrac{5}{4}\right)^4\)
Câu 3
1, với y-45=0 ta có y=45
=>2x+37=45-45+45-45=0(vô lý)
với y-45>0 ta có |y-45|=y-45
=>2x-37=2y-90
37+90=2y-2x
=>2y-2x=127 là số lẻ
\(\begin{cases}
2^x=1\\
2y=128
\end{cases}\)=>\(\begin{cases}
x=0\\
y=64
\end{cases}\)
2,
Giả sử b2−4acb2-4ac là số chính phương
Đặt b2−4ac=r2b2-4ac=r2
Xét 4a.¯¯¯¯¯abc4a.abc¯
=4a.(100a+10b+c)=4a.(100a+10b+c)
=400a2+40b+4ac=400a2+40b+4ac
=400a2+40b+b2−r2=400a2+40b+b2-r2
=(20a+b)2−r2=(20a+b)2-r2
=(20a+b− r)(20a+b+r)=(20a+b- r)(20a+b+r)
⇒4a=(20a+b−r)(20a+b+r)¯¯¯¯¯abc⇒4a=(20a+b-r)(20a+b+r)abc¯
Vì a∈Z⇒4a∈Za∈ℤ⇒4a∈ℤ
⇒(20a+b−r)(20a+b+r)⋮¯¯¯¯¯abc⇒(20a+b-r)(20a+b+r)⋮abc¯
Mà ¯¯¯¯¯abcabc¯ là số nguyên tố và 20a+b+r>20a+b−r20a+b+r>20a+b-r
⇒20a+b+r=¯¯¯¯¯abc;20a+b−r=1⇒20a+b+r=abc¯;20a+b-r=1
Ta có : 20a+b+r=100a+10b+c20a+b+r=100a+10b+c
⇒80a+9b+c=r⇒80a+9b+c=r (Vô lý do r<9b+cr<9b+c)
⇒⇒ Điều giả sử là sai
⇒đpcm
