Tìm y: (y-0,5)^4+(y+0,5)^4=1
(y-2)^5+(y+2)^5=242y
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a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
a: \(\Leftrightarrow y\cdot2.4=\dfrac{6}{5}\cdot\dfrac{2}{5}=\dfrac{12}{25}\)
hay y=1/5
b: 5/4:y=1/2
nên y=5/4:1/2=5/2
1)\(y\times7:5+4\times8=134\)
\(\Leftrightarrow y\times7:5+32=134\)
\(\Leftrightarrow y\times7:5=102\)
\(\Leftrightarrow y\times7=510\)
\(\Leftrightarrow y=72,86\)
2) \(\dfrac{1}{4}:0,25-\dfrac{1}{8}:0,125+\dfrac{1}{2}:0,5-\dfrac{1}{10}\)
\(=0,25:0,25-0,125:0,125+0,5:0,5-\dfrac{1}{10}\)
\(=1-1+1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
\(7,75-\left(0,5\times y\div5-6,2\right)=5\)
\(0,5\times y\div5-6,2=7,75-5=2,75\)
\(0,5\div5\times y-6,2=2,75\)
\(0,1\times y=2,75+6,2=8,95\)
\(\dfrac{1}{10}y=8,95\)
\(y=8,95\times10=89,5\)
\(y\div6\times7,2+1,3\times y+y\div2+15=19,95\)
\(1,2\times y+1,3\times y+0,5y=19,95-15=4,95\)
\(y\left(1,2+1,3+0,5\right)=4,95\)
\(2y=4,95\)
\(y=4,95\div2=2,475\)
\(1,\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{3x+y}{9+5}=\dfrac{28}{14}=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\\ 2,\\ a,a=2\Rightarrow y=f\left(x\right)=2x\\ b,f\left(-0,5\right)=2\left(-0,5\right)=-1\\ f\left(\dfrac{3}{4}\right)=2\cdot\dfrac{3}{4}=\dfrac{3}{2}\\ c,\text{Thay }x=-4;y=2\Rightarrow-4a=2\Rightarrow a=-\dfrac{1}{2}\)
Ta có: x/y=3/5 ⇒ x/3=y/5
Theo tính chất của dãy tỉ số bằng nhau ta có:x/3=y/5=3x/3.3=y/5= 3x+y9/y9+5=28/14=2
Do đó:
x/3=2 ⇒x=2.3=6
y/5=2 ⇒y=2.5=10
Vậy x=6 và y=10.
Bài 2:
a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
hay \(x=\dfrac{2}{7}\)
b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow x^3=-8\)
hay x=-2
Bài 1:
a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)
\(=xy^2-x^2y^2+x^2y^2-xy^2+xy\)
\(=xy\)
=1
b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)
\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)
\(=x^2-y^2\)
\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)
(y - 0,5)4 + (y + 0,5)4 = 1
<=> (y - 0,5)4 + (y - 0,5 + 1)4 = 1
Đặt y - 0,5 = a
<=> a4 + (a + 1)4 = 1
<=> a4 + a4 + 4a3 + 6a2 + 4a + 1 = 1
<=> 2a4 + 4a3 + 6a2 + 4a = 0
<=> 2a(a3 + 2a2 + 3a + 2) = 0
<=> a(a3 + a2 + a2 + a + 2a + 2) = 0
<=> a(a + 1)(a2 + a + 2) = 0
<=> a(a + 1) = 0 (vì a2 + a + 2 = (a2 + a + 1/4) + 7/4 = (a + 1/2)2 + 7/4 > 0)
<=> \(\orbr{\begin{cases}a=0\\a+1=0\end{cases}}\) <=> \(\orbr{\begin{cases}a=0\\a=-1\end{cases}}\)
Với a = 0 => y - 0,5 = 0 <=> y = 0,5
Với a = -1 => y - 0,5 = -1 <=> y = -0,5
Vậy S = {0,5; -0,5}