Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
cmt gi may co bai tu di ma lam bat ng ta lam cho may chep ha
\(2x+\frac{-1}{2}=\frac{-2}{3}\)
<=>\(2x=\frac{-2}{3}-\frac{-1}{2}=\frac{-1}{6}\)
<=>\(x=\frac{-1}{6}:2=-\frac{1}{2}\)
Vây \(x=-\frac{1}{2}\)
\(0.75-\left(-2x\right)=\frac{4}{5}\)
<=>\(\frac{3}{5}+2x=\frac{4}{5}\)
<=>\(2x=\frac{4}{5}-\frac{3}{5}=\frac{1}{5}\)
<=>\(x=\frac{1}{5}:2=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
\(\left(2x+5\right)\left(1-x\right)=0\)
<=>\(2x+5=0\)hoặc \(1-x=0\)
<=>\(x=\frac{5}{2}\)hoặc \(x=1\)
Vậy \(x=\frac{5}{2}\)hoặc \(x=1\)
Bài làm:
a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
+ Nếu x = 6
\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)
+ Nếu x = 4
\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)
Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)
b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{4}{3}\)
Thay vào ta được:
\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)
\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)
\(\Leftrightarrow14y=-4\)
\(\Rightarrow y=-\frac{2}{7}\)
Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)
1/4x+2/3x+4/3=0
11/12x=0-4/3
11/12x=-4/3
x=-4/3:11/12
x=-48/33
\(A=-\left|2x-1\right|\)
Do \(-\left|2x-1\right|\le0\)
\(\Rightarrow Max\)\(A=-0=0\)
Vậy Max A=0 khi x=\(\frac{1}{2}\)
\(B=3-\left|2x-1\right|\)
Do \(\left|2x-1\right|\ge0\)
\(\Rightarrow Max\)\(B=3-0=3\)
Vậy \(Max\)\(B=3\)\(Khi\)\(x=\frac{1}{2}\)
\(C=-\left|2x-1\right|+1\)
Do \(-\left|2x-1\right|\le0\)
\(\Rightarrow Max\)\(C=0+1=1\)
Vậy \(Max\)\(C=1\)\(khi\)\(x=\frac{1}{2}\)
\(2^{24}=(2^3)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
vì \(8^8< 9^8\Rightarrow2^{24}< 3^{16}\)
\(2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
\(8< 9\)
\(\Rightarrow8^9< 9^9\)
\(\Rightarrow2^{24}< 3^{16}\)
1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab
=>(a+b/)2ab-1/h=0
quy dong len ta co
(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0
=>ah+bh-ab-ab=0
=>a(h-b)-b(a-h)=0
=>a(h-b)=b(a-h)
=>a/b=(a-h)(h-b)
Hình như bạn nhập sai đề bài rùi , thôi mik sửa theo cách mik thử
Nếu \(\left(\frac{1}{2}\right)^{2x}+1=\frac{1}{8}\)
Ta có: \(\left(\frac{1}{2}\right)^{2x}=-\frac{7}{8}\)
mà \(\left(\frac{1}{2}\right)^{2x}\ge0\forall x;-\frac{7}{8}< 0\)
\(\Rightarrow2x\in\varnothing\Rightarrow x\in\varnothing\)