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a) =-3/5 x (4/7 + 2/7 + 1) = -3/5 x 13/7 = -39/35
b) =8/11 x 5/19 + 7/19 x 8/11 - 8/11 x 1/15 = 8/11 x ( 5/19 + 7/19 - 1/15 ) = 8/11 x 161/285 = 1288/3235
**** cho mink nha
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)\(\Rightarrow\frac{a+b}{c}-1=\frac{b+c}{a}-1=\frac{c+a}{b}-1\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=\frac{2\left(a+b+c\right)}{a+b+c}\)(1)
Ta có: \(M=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\)
TH1: Nếu \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Rightarrow M=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=\frac{-abc}{abc}=-1\)
TH2: Nếu \(a+b+c\ne0\)\(\Rightarrow\)Biểu thức (1) bằng 2
\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)\(\Rightarrow M=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=\frac{8abc}{abc}=8\)
Vậy \(M=-1\)hoặc \(M=8\)
a) Ta có 3/7 < 3,5/7 = 1/2 = 7,5/15 <11/15
Vậy 3/7 < 11/15
b) -11/6 < -1 < -8/9
Ai đủ điểm hỏi đáp đi ngang qua tk giùm với :(
Bài 1:
\(\frac{x}{-8}=\frac{-18}{x}\)
\(\Rightarrow x^2=144\)
\(\Rightarrow x=\pm12\)
Vậy \(x=\pm12\)
Bài 3:
Giải:
Ta có: \(\frac{a}{b}=\frac{2,1}{2,7}\Rightarrow\frac{a}{2,1}=\frac{b}{2,7}\Rightarrow\frac{a}{21}=\frac{b}{27}\Rightarrow\frac{a}{7}=\frac{b}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{7}=\frac{b}{9}=\frac{5a}{35}=\frac{4b}{36}=\frac{5a-4b}{35-36}=\frac{-1}{-1}=1\)
+) \(\frac{a}{7}=1\Rightarrow a=7\)
+) \(\frac{b}{9}=1\Rightarrow b=9\)
\(\Rightarrow\left(a-b\right)^2=\left(7-9\right)^2=\left(-2\right)^2=4\)
Vậy \(\left(a-b\right)^2=4\)
Bài 4:
Giải:
Ta có: \(\frac{a}{b}=\frac{9,6}{12,8}\Rightarrow\frac{a}{9,6}=\frac{b}{12,8}\Rightarrow\frac{a}{96}=\frac{b}{128}\Rightarrow\frac{a}{3}=\frac{b}{4}\)
Đặt \(\frac{a}{3}=\frac{b}{4}=k\)
\(\Rightarrow a=3k,b=4k\)
Mà \(a^2+b^2=25\)
\(\Rightarrow\left(3k\right)^2+\left(4k\right)^2=25\)
\(\Rightarrow9.k^2+16.k^2=25\)
\(\Rightarrow25k^2=25\)
\(\Rightarrow k^2=1\)
\(\Rightarrow k=\pm1\)
+) \(k=1\Rightarrow a=3;b=4\)
+) \(k=-1\Rightarrow a=-3;b=-4\)
\(\Rightarrow\left|a+b\right|=\left|3+4\right|=\left|-3+-4\right|=7\)
Vậy \(\left|a+b\right|=7\)
Áp dụng BĐT
\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)Ta có:
\(\left|2x-7\right|+\left|2x+1\right|=\left|2x-7\right|+\left|-2x-1\right|\ge\left|2x-7+\left(-2x-1\right)\right|=8\)
Mà \(\left|2x-7\right|+\left|2x+1\right|\ge\)8 nên không có số nguyên x nào thỏa mãn đề ra
Ta có :
\(\frac{8}{11}+\frac{a}{b}=\frac{8}{11}.\frac{a}{b}\)
\(\Rightarrow\frac{a}{b}=\frac{8}{11}.\frac{a}{b}-\frac{8}{11}\)
\(\Rightarrow\frac{a}{b}=\frac{8}{11}.\left(\frac{a}{b}-1\right)\)
\(\Rightarrow\frac{a}{b}-1=\frac{8}{11}.\left(\frac{a}{b}-1\right)-1\)
\(\Rightarrow1=\frac{8}{11}.\left(\frac{a}{b}-1\right)-\left(\frac{a}{b}-1\right)\)
\(\Rightarrow\left(\frac{8}{11}-1\right).\left(\frac{a}{b}-1\right)=1\)
\(\Rightarrow\frac{-3}{11}.\left(\frac{a}{b}-1\right)=1\)
\(\Rightarrow\frac{a}{b}-1=1:\frac{-3}{11}=\frac{-11}{3}\)
\(\Rightarrow\frac{a}{b}=\frac{-11}{3}+1=\frac{-8}{3}\)
\(\frac{8}{11}+\frac{a}{b}=\frac{8}{11}.\frac{a}{b}\)
\(\Leftrightarrow\frac{8}{11}+\frac{a}{b}=\frac{8a}{11b}\)
\(\Leftrightarrow\frac{a}{b}-\frac{8a}{11b}=-\frac{8}{11}\)
\(\Leftrightarrow\frac{3a}{11b}=-\frac{8}{11}\)
\(\Leftrightarrow3a=-8b\)
\(\Leftrightarrow\frac{a}{b}=-\frac{8}{3}\)