Tìm các STN x va y
a.2x+1.3y=12x
b.10x:5y=20y
c.2x+2y=2x+y
d.2x-2y=258
Giúp mk vs mn ơi, mk cảm ơn trc
K
Khách
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3 tháng 12 2021
a: \(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
LT
1
KN
9 tháng 8 2019
a) \(A=7x^2-2x+1=7\left(x^2-\frac{2}{7}x+\frac{1}{7}\right)\)
\(=7\left(x^2+\frac{2}{7}x+\frac{1}{49}+\frac{6}{49}\right)\)
\(=7\left[\left(x+\frac{1}{7}\right)^2+\frac{6}{49}\right]=7\left(x+\frac{1}{7}\right)^2+\frac{6}{7}\ge\frac{6}{7}\)
Vậy \(A_{min}=\frac{6}{7}\Leftrightarrow x=\frac{-1}{7}\)
17 tháng 3 2019
ĐKXĐ: \(2x-y-1\ge0;x+2y\ge0\)
Đặt \(\sqrt{2x-y-1}=a;\sqrt{x+2y}=b\left(a,b\ge0\right)\). Khi đó ta có:
\(\left(2b^2-1\right)a=\left(2a^2-1\right)b\Leftrightarrow\left(a-b\right)\left(2ab+1\right)=0\)
\(\Leftrightarrow a=b\) hoặc \(2ab+1=0\)(loại vì \(a,b\ge0\))
Suy ra: \(\sqrt{2x-y-1}=\sqrt{x+2y}\Leftrightarrow x=3y+1\)
Pt đầu tiên trở thành: \(\left(3y+1\right)^2-5y^2-8y=3\)
\(\Leftrightarrow\left(y-1\right)\left(2y+1\right)=0\Leftrightarrow\orbr{\begin{cases}y=1\\y=-\frac{1}{2}\end{cases}}\)
+) Với \(y=1\Rightarrow x=4\Rightarrow\left(x;y\right)=\left(4;1\right)\)(tm)
+) Với \(y=-\frac{1}{2}\Rightarrow x=-\frac{1}{2}\Rightarrow\left(x;y\right)=\left(-\frac{1}{2};-\frac{1}{2}\right)\) (loại)
Vậy hpt có nghiệm duy nhất \(\left(x;y\right)=\left(4;1\right).\)
1 tháng 10 2017
\(2x^2+2y^2-4xy+2x-2y+4\)
\(=2\left(x-y\right)^2+2\left(x-y\right)+4\)
\(=2\left[\left(x-y\right)^2+2\left(x-y\right)\frac{1}{2}+\frac{1}{4}\right]+\frac{7}{2}\)
\(=2\left(x-y+\frac{1}{2}\right)^2+\frac{7}{2}\)
\(\Rightarrow A\ge\frac{7}{2}\)
Dấu = bn tự tính nhé
RW
2
T
2 tháng 6 2017
\(\left(2x-1\right)^3-8\left(x-1\right)\left(x^2+x+1\right)+12x^2=2x+1\)
\(\Leftrightarrow8x^3-12x^2+6x-1-8\left(x^3-1\right)+12x^2-2x-1=0\)
\(\Leftrightarrow4x+6=0\)
\(\Leftrightarrow2\left(2x+3\right)=0\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\frac{-3}{2}\)
MC
1
18 tháng 7 2020
Ta có: \(-2x\left(x+5\right)+\left(2x^2+4\right)+10x\)
\(=-2x^2-10x+2x^2+4+10x\)
=4
9 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
29 tháng 6 2019
I 2x-3 I = I x+1 I
2x-3 = x+1
x+1 - 2x+3=0
x (1-2) +1+3=0
-1x +4 =0
-1x = 0-4
-1x =-4
x = -4 : -1
x =4
29 tháng 6 2019
Trả lời:
\(\left|2x-3\right|=\left|x+1\right|\)
\(\Rightarrow2x-3=x+1\) hoặc \(2x-3=-\left(x+1\right)\)
TH1: \(2x-3=x+1\)
\(2x-x=1+3\)
\(x=4\)
TH2: \(2x-3=-\left(x+1\right)\)
\(2x-3=-x-1\)
\(2x+x=-1+3\)
\(3x=2\)
\(x=\frac{2}{3}\)
Vậy \(x=4;x=\frac{2}{3}\)
\(2^{x+1}.3^y=12^x\)
\(\Rightarrow2^{x+1}.3^y=3^x.4^x\)
\(\Rightarrow2^{x+1}.3^y=3^x.2^{2x}\)
\(\Rightarrow\orbr{\begin{cases}2^{x+1}=2^{2x}\\3^y=3^x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+1=2x\\y=x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\\text{Vì y = x}\Rightarrow y=1\end{cases}}\)