1/6x5/15+1/6x3/15
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\(P\left(x\right)=5^6-6.5^5+6.5^4-6.5^3+6.5^2-6.5+1=5^6-6\left(5^5-5^4-5^3-5^2-5\right)+1=1556\)
\(x=\)\(\frac{-30}{11}nha\)\(bạn\)\(!!!!!!!!\)
\(tk\)\(mik\)\(nha\)\(!!!!!!!!!!!!!\)
(6x5:X)+15=4
30:X=4-15=-11
=>X=30:(-11)=-\(\frac{30}{11}\)
hihi, chúc bn hk tốt nha >,<
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
\(\dfrac{3}{4}\times\dfrac{5}{7}=\dfrac{15}{28};\dfrac{5}{8}\times\dfrac{4}{15}=\dfrac{20}{120}=\dfrac{1}{6};\dfrac{7}{12}\times\dfrac{4}{9}=\dfrac{28}{108}=\dfrac{7}{27};\)
\(\dfrac{1}{6}\times\dfrac{3}{5}=\dfrac{3}{30}=\dfrac{1}{10};\dfrac{12}{21}\times\dfrac{23}{8}=\dfrac{276}{168}=\dfrac{23}{14};\dfrac{13}{4}\times\dfrac{5}{39}=\dfrac{65}{156}=\dfrac{5}{12}\)
\(\dfrac{7}{42}\times\dfrac{13}{21}=\dfrac{91}{882}=\dfrac{13}{126};\dfrac{3}{16}\times\dfrac{4}{15}=\dfrac{12}{240}=\dfrac{1}{20}\)
a) 11/12 + 5/6 x 3/2 - 2/3
= 11/12 + 15/12 - 2/3
= 26/12 - 2/3
= 26/12 - 8/12
= 18 /12 = 3/2
b) 4/9 : 4/6 x 3/4 +5/3
= 4/9 x 6/4 x 3/4 + 5/3
= 2/3 x 3/4 + 5/3
= 6/12 + 5/3
= 6/12 + 20/12
= 26/12 = 13/6
c) 5/7 : 11/14 : 15/22 x 3/5
= 5/7 x 14/11 x 22/15 x 3/5
= 10/11 x 22/15 x 3/5
= 4/3 x 3/5 = 20/15 = 4/3
a. \(\dfrac{11}{12}+\dfrac{5}{6}\)x\(\dfrac{3}{2}-\dfrac{2}{3}\) \(=\)\(\dfrac{11}{12}+\dfrac{5}{4}-\dfrac{2}{3}=\dfrac{11}{12}+\dfrac{15}{12}-\dfrac{8}{12}=\dfrac{11+15-8}{12}=\dfrac{18}{12}=\dfrac{3}{2}\)
b. \(\dfrac{4}{9}:\dfrac{4}{6}\)x\(\dfrac{3}{4}+\dfrac{5}{3}\) \(=\dfrac{4}{9}:\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{9}+\dfrac{5}{3}=\dfrac{8}{9}+\dfrac{15}{9}=\dfrac{8+15}{9}=\dfrac{23}{9}\)
c. \(\dfrac{5}{7}:\dfrac{11}{14}:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{10}{14}:\) \(\dfrac{11}{14}:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{10}{11}\)\(:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{20}{22}:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{4}{3}\)x\(\dfrac{3}{5}=\dfrac{4}{5}\)
a.
\(A=6\left(x^3+2^3\right)-6x^3-2\\ =6x^3+48-6x^3-2\\ =46\)
Vậy biểu thức trên không phụ thuộc vào giá trị x.
b.
\(B=2\left(\left(3x\right)^3+1\right)-54x^3\\ =2\left(27x^3+1\right)-54x^3\\ =54x^3+2-54x^3\\ =2\)
Vậy biểu thức trên không phụ thuộc vào giá trị x.
Câu 4:
D=(x+1)(x+3)(x+5)(x+7)+15
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+105+15\)
\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+120\)
\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x+2\right)\left(x+6\right)x^2+8x+10\)
Câu 2:
b: \(4x^2-12x+9\)
\(=\left(2x\right)^2-2\cdot2x\cdot3+3^2\)
\(=\left(2x-3\right)^2\)
Câu 1:
a: \(4x^2-9y^2=\left(2x\right)^2-\left(3y\right)^2=\left(2x-3y\right)\left(2x+3y\right)\)
b: \(\left(3x+y\right)^3=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot y+3\cdot3x\cdot y^2+y^3\)
\(=27x^3+27x^2y+9xy^2+y^3\)
\(\frac{1}{6} \times \frac{5}{15} + \frac{1}{6} \times \frac{3}{15}\)
\(\frac{1 \cdot 5}{6 \cdot 15} = \frac{5}{90} , \frac{1 \cdot 3}{6 \cdot 15} = \frac{3}{90}\)
\(\frac{5}{90} + \frac{3}{90} = \frac{8}{90}\)
\(\frac{8}{90} = \frac{4}{45}\)
\(\frac16\times\frac{5}{15}+\frac16\times\frac{3}{15}\)
\(=\frac16\times\left(\frac{5}{15}+\frac{3}{15}\right)\)
\(=\frac16\times\frac{8}{15}\)
\(=\frac{8}{90}=\frac{4}{45}\)