tìm x thuộc Z
a) 3x-4=-2x+5
b) 3(-4-2x) =3x+6
c) /3x-2/=2x-1
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(x+2\right)^2-\left(x+3\right)\left(x-1\right)+15\)
\(A=x^2+4x+4-\left(x^2-x+3x-3\right)+15\)
\(A=\left(x^2-x^2\right)+\left(4x+x-3x\right)+\left(15+3+4\right)\)
\(A=2x+22\)
______________________
\(B=\left(x+1\right)\left(x-1\right)-\left(x+4\right)^2-6\)
\(B=\left(x^2-1\right)-\left(x^2+8x+16\right)-6\)
\(B=\left(x^2-x^2\right)-8x-\left(1+16+6\right)\)
\(B=-8x-23\)
_________________
\(C=\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2\)
\(C=\left[\left(3x\right)^2-2^2\right]-\left(9x^2-6x+1\right)\)
\(C=\left(9x^2-9x^2\right)+6x-\left(4+1\right)\)
\(C=6x-5\)
a) Rút gọn biểu thức A = (x + 2)2 - (x + 3)(x - 1) + 15:
Bắt đầu bằng việc mở ngoặc:
A = (x^2 + 4x + 4) - (x^2 + 2x - 3x - 3) + 15
Tiếp theo, kết hợp các thành phần tương tự:
A = x^2 + 4x + 4 - x^2 - 2x + 3x + 3 + 15
Tiếp tục đơn giản hóa:
A = x^2 - x^2 + 4x - 2x + 3x + 4 + 3 + 15
Kết quả cuối cùng:
A = 5x + 19
b) Rút gọn biểu thức B = (x - 1)(x + 1) - (x + 4)2 - 6:
Bắt đầu bằng việc mở ngoặc:
B = (x^2 - 1) - (x^2 + 4x + 4) - 6
Tiếp theo, kết hợp các thành phần tương tự:
B = x^2 - 1 - x^2 - 4x - 4 - 6
Tiếp tục đơn giản hóa:
B = x^2 - x^2 - 4x - 4 - 6 - 1
Kết quả cuối cùng:
B = -4x - 11
c) Rút gọn biểu thức C = (3x - 2)(3x + 2) - (3x - 1)2:
Bắt đầu bằng việc mở ngoặc:
C = (9x^2 - 4) - (9x^2 - 6x + 1)
Tiếp theo, kết hợp các thành phần tương tự:
C = 9x^2 - 4 - 9x^2 + 6x - 1
Tiếp tục đơn giản hóa:
C = 9x^2 - 9x^2 + 6x - 4 - 1
Kết quả cuối cùng:
C = 6x - 5
a) \(2x-4< 5\)
\(\Leftrightarrow\) \(2x< 4+5\)
\(\Leftrightarrow\) \(x< 4,5\)
b) \(4-3x\ge6\)
\(\Leftrightarrow\) \(-3x\ge-4+6\)
\(\Leftrightarrow-3x\ge2\)
\(\Leftrightarrow\) \(x\le-0,6\)
c) \(3x-7< 5x-2\)
\(\Leftrightarrow\) \(3x-5x< 7-2\)
\(\Leftrightarrow\) \(-2,5x< 5\)
\(\Leftrightarrow x>-2,5\)
Bài 1:
a: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{2;0;4;-2\right\}\)
a) PT \(\Leftrightarrow x^2-x-x^2+2x=5\) \(\Rightarrow x=5\)
Vậy ...
b) PT \(\Leftrightarrow8x=16\) \(\Rightarrow x=2\)
Vậy ...
a: Ta có: \(x\left(x-1\right)-x^2+2x=5\)
\(\Leftrightarrow x^2-x-x^2+2x=5\)
hay x=5
b: Ta có: \(2x\left(3x+4\right)-6x^2=16\)
\(\Leftrightarrow6x^2+8x-6x^2=16\)
\(\Leftrightarrow8x=16\)
hay x=2
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)
\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)
\(\Rightarrow x\left(6x-2-15-6x\right)\)
\(\Rightarrow-16x=0\)
\(\Rightarrow x=0\)
d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)
\(\Rightarrow9x^2-4-4x+4=0\)
\(\Rightarrow9x^2-4x=0\)
\(\Rightarrow x\left(9x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
\(\left(-3x+2\right)-\left(5-3x\right)=-3\)
\(\Rightarrow-3x+2-5+3x=-3\)
\(\Rightarrow-3x+3x=-3+5-2\)
\(\Rightarrow0x=0\Rightarrow x\in Z\)
\(3+x-\left(3x-1\right)=6-2x\)
\(\Rightarrow3+x-3x+1=6-2x\)
\(\Rightarrow x-3x+2x=6-1-3\)
\(\Rightarrow0x=2\left(loại\right)\)
\(\left(x-5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)
\(7x\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
\(\left(3x-1\right)2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)
1.
a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)
b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)
2.
a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)
b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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
a) 3x - 4 = -2x + 5
<=> 3x = -3x + 9
<=> 3x + 2x = 9
<=> 5x = 9
<=> x = 9/5
\(\text{a) 3x-4=-2x+5}\)
\(3x-4=-2x+5\)
\(3x+2x=5+4\)
\(5x=9\)
\(\Rightarrow x=\frac{9}{5}\)
\(\text{b) 3(-4-2x) =3x+6}\)
\(-12-6x=3x+6\)
\(-6x-3x=6+12\)
\(-9x=18\)
\(\Rightarrow x=-2\)
\(\text{c) /3x-2/=2x-1}\)
\(\Rightarrow\orbr{\begin{cases}3x-2=2x-1\\3x-2=-\left(2x-1\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}3x-2=2x-1\\3x-2=-2x+1\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2x=-1+2\\3x+2x=1+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\5x=3\end{cases}}}\)\(\Leftrightarrow x\in\left\{1;\frac{3}{5}\right\}\)
học tốt