Y x 3/5 + y x 2/7 = 31/70
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NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 2 2022
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=3\\21\left(x^5+y^5\right)=31.3\left(x^3+y^3\right)\end{matrix}\right.\)
\(\Rightarrow31\left(x^2+xy+y^2\right)\left(x^3+y^3\right)=21\left(x^5+y^5\right)\)
\(\Rightarrow10x^5+31x^4y+31x^3y^2+31x^2y^3+31xy^4+10y^5=0\)
\(\Rightarrow\left(x+y\right)\left(x+2y\right)\left(2x+y\right)\left(5x^2-2xy+5y^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=-x\\y=-\dfrac{1}{2}x\\y=-2x\\y=x=0\end{matrix}\right.\) \(\Rightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2021
\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(1;1\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2021
2.
ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)
\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)
\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)
\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\3-x+2\sqrt{x\left(3-2x\right)}=4\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)
\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)
\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
9 tháng 2 2021
l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
9 tháng 2 2021
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8
ND
0
TA
19 tháng 1 2018
a/ -3x=-60+30
=> -3x=-30
=> x=(-30):(-3)
=> x=10
Ban oi phai x,y∈Z thi moi lam dc
b/x.y=7
Vi x,y∈U(7)=(1,-1,7,-7)
Với x=1 => y=7 y=1=>x=7
x=-1 =>x=-7 y=-1=>x=-7
x=7 => x=1 y=7=>x=7
x=-7 => x=-1 y=-7=>x=-1
c/(x-1)(y+2)=-7
x-1 va y+2∈U(-7)=(1,-1,7,-7)
Bạn giải giống câu b/ nhưng x,y trái đâu nhé
d/2x-3=5x-7
=>(-3)+7=5x-2x
=>4=3x
=>x=4/3
e/(-2)2-8x=20
=>4-8x=20
=>4-20=8x
=>-16=8x
=>x=-2
f/17-(-31)+3x=2x-(-50)
=>17+31+3x=2x+50
=>49+3x=2x+50
=> 3x-2x=50-49
=> x=1
g/2y2-16=34
=>2y2=34+16
=>2y2=50
=>y2=50:2
=>y2=25=52=(-5)2
=>y=5 va y=-5
PN
19 tháng 1 2018
a) \(\left(-3\right)x-30=-60\)
\(\Leftrightarrow\left(-3\right)x=-60+30\)
\(\Leftrightarrow\left(-3\right)x=-30\)
\(\Leftrightarrow x=\left(-30\right):\left(-3\right)\)
\(\Leftrightarrow x=10\)
b) \(xy=7\)
\(\Rightarrow xy=1.7=7.1=\left(-1\right).\left(-7\right)=\left(-7\right).\left(-1\right)\)
Ta có bảng sau:
| \(x\) | \(1\) | \(7\) | \(-1\) | \(-7\) |
| \(y\) | \(7\) | \(1\) | \(-7\) | \(-1\) |
KL: Các cặp số (x; y)...
c) \(\left(x-1\right)\left(y+2\right)=-5\)
\(\Rightarrow\left(x-1\right)\left(y+2\right)=1.\left(-5\right)=\left(-5\right).1=\left(-1\right).5=5.\left(-1\right)\)
Ta có bảng sau:
| \(x-1\) | \(1\) | \(-5\) | \(5\) | \(-1\) |
| \(y+2\) | \(-5\) | \(1\) | \(-1\) | \(5\) |
| \(x\) | \(2\) | \(-4\) | \(6\) | \(0\) |
| \(y\) | \(-7\) | \(-1\) | \(-3\) | \(3\) |
KL: Các cặp số (x; y)...
d) \(2x-3=5x-7\)
\(\Leftrightarrow2x-5x=-7+3\)
\(\Leftrightarrow-3x=-4\)
\(\Leftrightarrow x=\dfrac{4}{3}\)
e) \(\left(-2\right)^2-8x=20\)
\(\Leftrightarrow4-8x=20\)
\(\Leftrightarrow8x=4-20\)
\(\Leftrightarrow8x=-16\)
\(\Leftrightarrow x=-2\)
f) \(17-\left(-31\right)+3x=2x-\left(-50\right)\)
\(\Leftrightarrow17+31+3x=2x+50\)
\(\Leftrightarrow48+3x=2x+50\)
\(\Leftrightarrow3x-2x=50-48\)
\(\Leftrightarrow x=2\)
g) \(2y^2-16=34\)
\(\Leftrightarrow2y^2=34+16\)
\(\Leftrightarrow2y^2=50\)
\(\Leftrightarrow y^2=50:2=25\)
\(\Leftrightarrow y^2=5^2=\left(-5\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)
21 tháng 6 2022
Bài 3:
a: =>|x+7|=0
=>x+7=0
hay x=-7
b: =>|x-3|=4
=>x-3=4 hoặc x-3=-4
=>x=7 hoặc x=-1
c: =>x-5<=0
hay x<=5
\(y\times\frac{3}{5}+y\times\frac{2}{7}=\frac{31}{70}\)
\(\Leftrightarrow y\times\left(\frac{3}{5}+\frac{2}{7}\right)=\frac{31}{70}\)
\(\Leftrightarrow y\times\frac{31}{35}=\frac{31}{70}\)
\(\Leftrightarrow y=\frac{1}{2}\)
y x \(\frac{3}{5}\)+ y x \(\frac{2}{7}\)= \(\frac{31}{70}\)
y x \(\left(\frac{3}{5}+\frac{2}{7}\right)\)= \(\frac{31}{70}\)
y x \(\frac{31}{35}\) = \(\frac{31}{70}\)
y = \(\frac{31}{70}\): \(\frac{31}{35}\)
y = \(\frac{1}{2}\)