Tìm x; y, biết:
a) 2x+3=y2 (x;y thuộc N )
b) (5x-1)2013 =(5x-1)2015
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a) 2x.(3x2 – 5x + 3)
=2x3-10x2+6x
b(-2x-1).( x2 + 5x – 3 ) – (x-1)3
=-2x3 - 10x2 + 6x - x2 - 5x + 3 - x3 + 3x2 - 3x + 1
= -3x3 - 8x2 - 2x + 4
d) (6x5y2 – 9x4y3 + 15x3y4) : 3x3y2
=2x2-3xy+5y2
a, \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-3,2y-6\in Z\\x-3,2y-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\end{matrix}\right.\)
Ta có bảng:
x-3 | -1 | -5 | 1 | 5 |
2y-6 | -5 | -1 | 5 | 1 |
x | 2 | -2 | 4 | 8 |
y | \(\dfrac{1}{2}\left(loại\right)\) | \(\dfrac{5}{2}\left(loại\right)\) | \(\dfrac{11}{2}\left(loại\right)\) | \(\dfrac{7}{2}\left(loại\right)\) |
Vậy không có x,y thỏa mãn đề bài
b, tương tự câu a
\(c,xy-5x+2y=7\\ \Rightarrow x\left(y-5\right)+2y-10=-3\\ \Rightarrow x\left(y-5\right)+2\left(y-5\right)=-3\\ \Rightarrow\left(x+2\right)\left(y-5\right)=-3\)
Rồi làm tương tự câu a
\(d,xy-3x-4y=5\\ \Rightarrow x\left(y-3\right)-4y+12=17\\ \Rightarrow x\left(y-3\right)-4\left(y-3\right)=17\\ \Rightarrow\left(x-4\right)\left(y-3\right)=17\)
Rồi làm tương tự câu a
I don't now
sorry
...................
nha
b) \(\left(3x-2\right)\left(x+1\right)^2\left(3x+8\right)=-16\)
\(\Leftrightarrow\)\(\left(3x-2\right)\left(3x+3\right)^2\left(3x+8\right)+144=0\)
Đặt: \(3x+3=a\)pt trở thành:
\(\left(a-5\right)a^2\left(a+5\right)+144=0\)
\(\Leftrightarrow\)\(a^4-25a^2+144=0\)
\(\Leftrightarrow\)\(\left(a-4\right)\left(a-3\right)\left(a+3\right)\left(a+4\right)=0\)
đến đây bạn tìm a rồi tính x
c) \(\left(4x-5\right)\left(2x-3\right)\left(x-1\right)=9\)
\(\Leftrightarrow\)\(\left(4x-5\right)\left(4x-6\right)\left(4x-4\right)-72=0\)
Đặt \(4x-5=a\)pt trở thành:
\(a\left(a-1\right)\left(a+1\right)-72=0\)
\(\Leftrightarrow\)\(a^3-a-72=0\)
p/s: ktra lại đề
d) \(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
\(\Leftrightarrow\)\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)=0\)
\(\Leftrightarrow\)\(\left[\left(2x^2+x-2013\right)-2\left(x^2-5x-2012\right)\right]^2=0\)
\(\Leftrightarrow\)\(\left(11x+2011\right)^2=0\)
đến đây làm nốt
a) \(|4-2x|+|x-2|=3-x\) ( 1 )
+) Với : x ≥ 2 , ta có :
( 1 ) \(\Leftrightarrow2x-4+x-2=3-x\)
\(\Leftrightarrow4x=9\)
\(\Leftrightarrow x=\dfrac{9}{4}\left(TM\right)\)
+) Với : x < 2 , ta có :
( 1 ) \(\Leftrightarrow4-2x+2-x=3-x\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\dfrac{3}{2}\left(TM\right)\)
KL........
b) Vô nghiệm
c/ ĐKXĐ: \(x\ge3\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x-3}-\sqrt{x-2}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\left(\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}\right)-\left(\sqrt{\left(x-1\right)\left(x+3\right)}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}-\sqrt{x+3}=0\\\sqrt{x-1}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2< 3\left(ktm\right)\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
a) \(\Rightarrow\left(x-1\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b) \(\Rightarrow25x^2+10x+1-25x^2+9=30\)
\(\Rightarrow10x=20\Rightarrow x=2\)
a. x2 - 2x + 1 = 25
<=> x2 - 2x - 24 = 0
<=> x2 - 6x + 4x - 24 = 0
<=> x(x - 6) + 4(x - 6) = 0
<=> (x + 4)(x - 6) = 0
<=> \(\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)
b. (5x + 1)2 - (5x - 3)(5x + 3) = 30
<=> 25x2 + 10x + 1 - 25x2 + 9 = 30
<=> 25x2 - 25x2 + 10x = 30 - 1 - 9
<=> 10x = 20
<=> x = 2
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1