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Lời giải:
$\frac{3}{5}(x-\frac{2}{3})^3=-16\frac{1}{5}=\frac{-81}{5}$
$\Leftrightarrow (x-\frac{2}{3})^3=\frac{-81}{5}: \frac{3}{5}=-27=(-3)^3$
$\Leftrightarrow x-\frac{2}{3}=-3$
$\Leftrightarrow x=\frac{-7}{3}$
Thay x = 2; y = -2 vào biểu thức x(x2 - y)(x3 - 2y2)(x4 - 3y3)(x5 - 4y4) ta được :
x3 - 2y2 = 23 - 2 x (-2)2 = 8 - 8 = 0
Vậy giá trị biểu thức x(x2 - y)(x3 - 2y2)(x4 - 3y3)(x5 - 4y4) tại x = 2; y = -2 là 0
sau 25 phút nữa. ( tức là lúc 5h25p, cả kim phút và giờ đều chỉ số 5)
2m=2n+1024
\(\Leftrightarrow2^m-2^n=1024=2^{10}\)
\(\Leftrightarrow2^n\left(2^{m-n}-1\right)=2^{10}\)
Nếu \(m-n=0\) (vô lý)
Nếu \(m-n>0\)
\(\Rightarrow2^{m-n}-1\) lẻ mà 28 chẵn
\(\Rightarrow2^{m-n}-1=1\Rightarrow m=n+1\)
\(\Rightarrow2^n=2^8\Rightarrow n=8\Rightarrow m=n+1=8+1=9\)
Vậy n=8;m=9
Do K là trung điểm cạnh huyền BC nên AK là đường trung tuyến ứng với cạnh huyền. Suy ra KA = KB= KC.
Do KD = KA nên KA = KB = KC = KD, hay AD = BC.
Xét tam giác KAC có KA = KC nên nó là tam giác cân. Vậy thì \(\widehat{KCA}=\widehat{KAC}\)
Xét tam giác ABC và CDA có: AD = BC, AC chung, \(\widehat{KCA}=\widehat{KAC}\) nên \(\Delta ABC=\Delta CDA\left(c-g-c\right)\Rightarrow\widehat{DCA}=\widehat{BAC}=90^o\)
Hay \(DC⊥AC.\)
\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{8}\right)^2\)
\(=>\left(\dfrac{1}{2}\right)^n=\left[\left(\dfrac{1}{2}\right)^3\right]^2\)
\(=>\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^6\)
\(\Rightarrow n=6\)
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