5/8 + 0,2 : | -x/4 + 2/5 | = 3/2
help me please T o T
Những câu hỏi liên quan
| -2/3 x + 0,4 | . 5/3 - 3/2= 1/6
help me please T o T
x^2+1/x+x/x^2+1=-5/2
help me! thanks
tìm cực trị:y=(1-x)^3(3x-8)^2
help me!!!
B=\(\dfrac{4^5.9^4-2^{10}.9^5}{2^{10}.3^8+6^8.20}\)
Tính B?
HELP ME, PLEASE
\(B=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^{10}}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3^2\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1-9}{6}=\dfrac{-8}{6}=-\dfrac{4}{3}\)
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1, Tính
a, 7/8 - 2/3
b, 7/8 - 2/3 × 3/4
2, Tìm X
a, 5/8 - x = 5/9
b, 3/4 : x = 2/5
( please help me!!
\(1-\)\(a,\frac{7}{8}-\frac{2}{3}=\frac{21}{24}-\frac{16}{24}=\frac{5}{24}\)
\(b,\frac{7}{8}-\frac{2}{3}\times\frac{3}{4}=\frac{7}{8}-\frac{1}{2}=\frac{7}{8}-\frac{4}{8}=\frac{3}{8}\)
\(2-\)\(a,\frac{5}{8}-x=\frac{5}{9}\)
\(x=\frac{5}{8}-\frac{5}{9}\)
\(x=\frac{5}{72}\)
\(b,\frac{3}{4}:x=\frac{2}{5}\)
\(x=\frac{3}{4}:\frac{2}{5}\)
\(x=\frac{15}{8}\)
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1, Tính:
\(\frac{7}{8}\) - \(\frac{2}{3}\) = \(\frac{21}{24}\) - \(\frac{16}{24}\) = \(\frac{5}{24}\)
\(\frac{7}{8}\) - \(\frac{2}{3}\) x \(\frac{3}{4}\) = \(\frac{7}{8}\) - \(\frac{1}{2}\) = \(\frac{7}{8}\) - \(\frac{4}{8}\) = \(\frac{3}{8}\)
2, Tìm x
\(\frac{5}{8}\)- x = \(\frac{5}{9}\)
x = \(\frac{5}{8}\)- \(\frac{5}{9}\)
x = \(\frac{5}{72}\)
\(\frac{3}{4}\): x = \(\frac{2}{5}\)
x = \(\frac{3}{4}\) : \(\frac{2}{5}\)
x = \(\frac{15}{8}\)
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(x-1)^2 = (2x-3)^2
help me
\(\Rightarrow\left(x-1\right)^2-\left(2x-3\right)^2=0\\ \Rightarrow\left(x-1-2x+3\right)\left(x-1+2x-3\right)=0\\ \Rightarrow\left(2-x\right)\left(3x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right.\)
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Nguyễn Hoàng Minh thank you
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(2x-3)^2
(x+5)^2
help meee
Tim x: 3.2^x+2^x=16
(3x-2)^2+1/4=1/2
Thuc hien phep tinh: 5.(-0,2)^2+3/8.(9/2-2,5)^2-(-1/5)
Mn giup mk nha
Please
Phân tích đa thức thành nhân tử:
a) x^8 + x^4 + 1
b) x^10 + x^5 + 1
(((((( Help me , Please !! ))))))
a)\(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
b)\(x^{10}+x^5+1\)
\(=\left(x^{10}+x^9+x^8\right)-\left(x^9+x^8+x^7\right)+\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^8\left(x^2+x+1\right)-x^7\left(x^2+x+1\right)+x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)
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a) \(x^8+x^4+1\)
= \(x^8+2x^4-x^4+1\)
= \(\left(x^4+1\right)^2-x^4\)
= \(\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\)
= \(\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\)
= \(\left(x^4-x^2+1\right)\left[\left(x^2+1\right)^2-x^2\right]\)
= \(\left(x^4-x^2+1\right)\left(x^2+1-x^2\right)\left(x^2+1+x^2\right)\)
= \(\left(x^4-x^2+1\right)\left(2x^2+1\right)\)
b) \(x^{10}+x^5+1\)
= ( x10+x9+x8) - (x9+x8+x7) + (x7+x6+x5) - (x6+x5+x4) + (x5+x4+x3) - (x3+x2+x) + (x2+x+1)
= (x2+x+1)(x8 - x7+x5-x4+x3-x+1)
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