C1 lấy 50.2/100=1Hz

50×2/100=1Hz

a   \(2a>b;2a>0\Rightarrow2a+2a>b+0\Rightarrow4a>b\)

b   \(4a^2+b^2=5ab\Rightarrow4a^2+b^2-5ab=0\Rightarrow\left(4a^2-4ab\right)-\left(ab-b^2\right)=0\)

\(\Rightarrow4a\left(a-b\right)-b\left(a-b\right)=0\Rightarrow\left(4a-b\right)\left(a-b\right)=0\Rightarrow\hept{\begin{cases}4a-b=0\Rightarrow4a=b\\a-b=0\Rightarrow a=b\end{cases}}\)

c  \(20=4\cdot5>11\)mà \(2\cdot5=10>11\)đâu 

sai đề r

Viết rõ đề bài ra đc không ạ

đấy là phân số

Bài làm:

a) đkxđ: \(2a\ne\pm b\)

Ta có: \(M=\left(\frac{2}{2a-b}+\frac{6b}{b^2-4a^2}-\frac{4}{2a+b}\right)\div\left(\frac{1+4a^2+4b^2}{4a^2-b^2}\right)\)

\(M=\left[\frac{2\left(2a+b\right)-6b-4\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\right].\left(\frac{\left(2a-b\right)\left(2a+b\right)}{4a^2+4b^2+1}\right)\)

\(M=\frac{4a+2b-6b-8a+4b}{4a^2+4b^2+1}\)

\(M=\frac{-4a}{4a^2+4b^2+1}\)

b) +Nếu: \(a=\frac{1}{3}\)và \(b=2\)

Khi đó GT của M là: \(M=\frac{-4.\frac{1}{3}}{4.\frac{1}{3^2}+4.2^2+1}=-\frac{12}{157}\)

Viết rõ đề ra nhá

a) \(A=\left(\frac{2}{2a-b}+\frac{6b}{b^2-4a^2}-\frac{4}{2a+b}\right):\left(a+\frac{4a^2+b^2}{4a^2-b^2}\right)\)

\(=\left(\frac{2}{2a-b}+\frac{6b}{\left(b-2a\right)\left(b+2a\right)}-\frac{4}{2a+b}\right):\left(a+\frac{4a^2+b^2}{4a^2-b^2}\right)\)

\(=\left(\frac{-2\left(b+2a\right)}{\left(b-2a\right)\left(b+2a\right)}+\frac{6b}{\left(b-2a\right)\left(b+2a\right)}-\frac{4\left(b-2a\right)}{\left(2a+b\right)\left(b-2a\right)}\right):\left(\frac{a\left(4a^2-b^2\right)}{4a^2-b^2}+\frac{4a^2+b^2}{4a^2-b^2}\right)\)

\(=\frac{-2b-4a+6b-4b+8a}{\left(b-2a\right)\left(b+2a\right)}:\frac{4a^3-ab^2+4a^2+b^2}{4a^2-b^2}\)

\(=\frac{4a}{\left(b-2a\right)\left(b+2a\right)}.\frac{\left(2a-b\right)\left(2a+b\right)}{4a^3-ab^2+4a^2+b^2}\)

\(=\frac{-4a}{\left(2a-b\right)\left(b+2a\right)}.\frac{\left(2a-b\right)\left(2a+b\right)}{4a^3-ab^2+4a^2+b^2}\)

\(=.\frac{-4a}{4a^3-ab^2+4a^2+b^2}\)

b)  ĐKXĐ: \(\hept{\begin{cases}2a\ne b\\2a\ne-b\end{cases}}\)

Ta thấy \(a=\frac{1}{3};b=2\)thỏa mãn điều kiện \(\hept{\begin{cases}2a\ne b\\2a\ne-b\end{cases}}\)nên thay vào A ta được:

bạn thay vào tự tính nhé mà cái phần rút gọn bạn vừa làm vừa check giùm bài mik nhé =)) sợ sai 

Lớp 7 gì mà dễ ẹc :))

\(\frac{2a-b}{a+b}=\frac{2}{3}\)

\(\Leftrightarrow6a-3b=2a+2b\)

\(\Rightarrow4a=5b\)

\(\frac{b-c+a}{2a-b}=\frac{2}{3}\)

\(\Leftrightarrow4a-2b=3b-3c+3a\)

\(\Leftrightarrow a=5b-3c\)

\(\Leftrightarrow a-5b=-3c\)

\(\Leftrightarrow a-4a=-3c\)

\(\Leftrightarrow-3a=-3c\)

\(\Rightarrow a=c\)

Ta có : \(P=\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=8\)

8 nhé bn !

a) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{4a}{3b}=\frac{4c}{3d}\)

Áp dụng dãy tỉ số bằng nhau ta có :

\(\frac{4a}{3b}=\frac{4c}{3d}\Rightarrow\frac{4a-3b}{4a+3b}=\frac{4c-3d}{4c+3d}\Rightarrow\frac{4a-3d}{4c-3d}=\frac{4a+3b}{4c+3d}\)

b) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{3b}=\frac{2c}{3d}\)

Áp dụng dãy tỉ số bằng nhau ta có :

\(\frac{2a}{3b}=\frac{2c}{2d}\Rightarrow\frac{2a-3b}{2a+3b}=\frac{2c-3d}{2c+3d}\)

\(A=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{\left(2a-1\right)^2}{2a+1}\cdot\dfrac{1}{\left(2a-1\right)\left(2a+1\right)}\right)\cdot\left(\dfrac{4a\left(a+1\right)+1}{4a^2}\right)-\dfrac{1}{2a}\)

\(=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{2a-1}{\left(2a+1\right)^2}\right)\cdot\dfrac{4a^2+4a+1}{4a^2}-\dfrac{1}{2a}\)

\(=\dfrac{-\left(2a-1\right)\left(2a+1\right)}{\left(2a+1\right)^2}\cdot\dfrac{\left(2a+1\right)^2}{4a^2}-\dfrac{1}{2a}\)

\(=\dfrac{-\left(4a^2-1\right)}{4a^2}-\dfrac{2a}{4a^2}\)

\(=\dfrac{-4a^2-2a+1}{4a^2}\)