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a: \(\Leftrightarrow9-x=6\)
hay x=3
e: \(\Leftrightarrow2^x=32\)
hay x=5
2:
a: x-1/4=7/12
=>x=7/12+3/12=10/12=5/6
b: 4x/3=-8/9
=>4x=-24/9=-8/3
=>x=-2/3
c: 3/2(x+5)=-12/5
=>x+5=-12/5:3/2=-12/5*2/3=-24/15=-8/5
=>x=-8/5-5=-33/5
d: x/4=16/x
=>x^2=64
=>x=8 hoặc x=-8
\(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{19.20}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{3}-\dfrac{1}{20}< \dfrac{1}{3}\)
\(B=\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{19.20}=2\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{19.20}\right)=2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)=2\left(\dfrac{1}{3}-\dfrac{1}{20}\right)=\dfrac{2}{3}-\dfrac{2}{20}< \dfrac{2}{3}\)
\(C=\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{18.20}=\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{18.20}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{18}-\dfrac{1}{20}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{4}-\dfrac{1}{40}< \dfrac{1}{4}\)
\(D=\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{18.20}=\dfrac{1}{2}\left(\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{18.20}\right)=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{18}-\dfrac{1}{20}\right)=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{20}\right)=\dfrac{1}{8}-\dfrac{1}{40}< \dfrac{1}{8}\)
Bài 9:
a: Bạn chỉ cần vẽ tam giác ABC vuông tại A có AB=3cm và AC=4cm là ra cái hình rồi
Số đo góc A thì chắc chắn là 90 độ rồi
\(B=\left(\dfrac{1}{9}-\dfrac{5}{7}\right)+\dfrac{3}{6}+\left(\dfrac{-12}{17}+\dfrac{-1}{2}\right)+\dfrac{5}{9}.\)
\(B=\dfrac{1}{9}-\dfrac{5}{7}+\dfrac{1}{2}-\dfrac{12}{17}-\dfrac{1}{2}+\dfrac{5}{9}.\)
\(B=\left(\dfrac{1}{9}+\dfrac{5}{9}\right)-\dfrac{5}{7}-\dfrac{12}{17}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right).\)
\(B=\dfrac{2}{3}-\dfrac{5}{7}-\dfrac{12}{17}=\dfrac{-269}{357}.\)
Bài 6:
\(B=\dfrac{1}{199}+1+\dfrac{2}{198}+1+\dfrac{3}{197}+1+...+\dfrac{198}{2}+1+1\)
\(=\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{2}+\dfrac{200}{200}\)
\(=200\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}\right)=200\cdot A\)
=>A/B=1/200
v: \(\dfrac{26}{-37}< 0\)
\(0< \dfrac{11}{19}\)
Do đó: \(\dfrac{26}{-37}< \dfrac{11}{19}\)
x: \(\dfrac{-7}{30}=\dfrac{-7\cdot2}{30\cdot2}=\dfrac{-14}{60}\)
\(\dfrac{13}{-20}=\dfrac{-13}{20}=\dfrac{-13\cdot3}{20\cdot3}=\dfrac{-39}{60}\)
mà -14>-39
nên \(-\dfrac{7}{30}< \dfrac{13}{-20}\)
y: \(-\dfrac{12}{16}=\dfrac{-12\cdot3}{16\cdot3}=\dfrac{-36}{48}\)
\(\dfrac{-16}{48}=\dfrac{-16}{48}\)
mà -36<-16
nên \(\dfrac{-12}{16}< -\dfrac{16}{48}\)
z: \(\dfrac{-18}{-72}=\dfrac{1}{4}>0\)
\(0>-\dfrac{5}{20}\)
Do đó: \(\dfrac{-18}{-72}>-\dfrac{5}{20}\)
z1: \(\dfrac{-36}{90}=\dfrac{-2}{5};\dfrac{-15}{25}=\dfrac{-3}{5}\)
mà -2>-3
nên \(\dfrac{-36}{90}>\dfrac{-15}{25}\)
z2: \(\dfrac{-32}{48}=\dfrac{-2}{3}=\dfrac{-4}{6}\)
\(\dfrac{-6}{12}=\dfrac{-1}{2}=\dfrac{-3}{6}\)
mà -4<-3
nên \(-\dfrac{32}{48}< -\dfrac{6}{12}\)
z3: \(\dfrac{-20}{-45}=\dfrac{4}{9}=\dfrac{40}{90}\)
\(\dfrac{35}{150}=\dfrac{7}{30}=\dfrac{21}{90}\)
mà 40>21
nên \(\dfrac{-20}{-45}>\dfrac{35}{150}\)
100.10.104=10.102.104=107
chúc bạn học tốt
10\(^7\)