\(2^{x+1}=1\)
\(\left(-2\right)^{x+2}=4\)
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8 tháng 1 2022
\(1,2\left(x-3\right)+1=2\left(x+1\right)-9\\ \Rightarrow2x-6+1=2x+2-9\\ \Rightarrow2x-5=2x-7\\ \Rightarrow-2=0\left(vô.lí\right)\)
\(2,\dfrac{5-x}{2}=\dfrac{3x-4}{6}\\ \Rightarrow30-6x=6x-8\\ \Rightarrow12x=38\\ \Rightarrow x=\dfrac{19}{6}\)
\(3,\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\\ \Rightarrow x^2-2x+1+x^2-4=2x^2-6x+x-3\\ \Rightarrow2x^2-2x-3=2x^2-5x-3\\ \Rightarrow3x=0\\ \Rightarrow x=0\)
\(4,\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\\ \Rightarrow x^2+5x-x-5-x^2-2x-x-2=1\\ \\ \Rightarrow x-7=1\\ \Rightarrow x=8\)
8 tháng 1 2022
\(5,\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\\ \Rightarrow\dfrac{6x-1}{15}-\dfrac{3x}{15}=\dfrac{10x}{15}\\ \Rightarrow6x-1-3x=10x\\ \Rightarrow3x-1=10x\\ \Rightarrow7x=-1\\ \Rightarrow x=\dfrac{-1}{7}\)
\(6,\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\\ \Rightarrow\dfrac{75\left(x-2\right)}{30}-\dfrac{10\left(x+5\right)}{30}=\dfrac{30}{30}-\dfrac{24\left(x-3\right)}{30}\\ \Rightarrow75\left(x-2\right)-10\left(x+5\right)=30-24\left(x-3\right)\\ \Rightarrow75x-150-10x-50=30-24x+72\\ \Rightarrow65x-200=102-24x\\ \Rightarrow89x=302\\ \Rightarrow x=\dfrac{320}{89}\)
22 tháng 2 2022
Phương pháp:
Đặt \(x+\dfrac{1}{x}=a\Rightarrow a^2=x^2+\dfrac{1}{x^2}+2\Leftrightarrow a^2-2=x^2+\dfrac{1}{x^2}\)
Thay vào pt
22 tháng 2 2022
\(x\ne0:đặt:x+\dfrac{1}{x}=t\)
\(pt\Leftrightarrow2t^2+4\left(t^2-2\right)^2-4\left(t^2-2\right)t^2=\left(x+4\right)^2\)
\(\Leftrightarrow2t^2+4\left(t^4-4t^2+4\right)-4\left(t^4-2t^2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow2t^2+4t^4-16t^2+16-4t^4+8t^2=\left(x+4\right)^2\)
\(\Leftrightarrow-6t^2+16=\left(x+4\right)^2\)
\(\Leftrightarrow-6\left(x^2+2+\dfrac{1}{x^2}\right)+16=x^2+8x+16\)
\(\Leftrightarrow-6x^2-\dfrac{6}{x^2}-x^2-8x-12=0\Leftrightarrow-6x^4-x^4-8x^3-12x^2-6=0\Leftrightarrow-7x^4-8x^3-12x^2-6=0\left(vô-nghiệm\right)\)
(bn xem lại đề)
DT
26 tháng 2 2022
đkxđ: x khác 0
\(\Leftrightarrow8.\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)+4\left(x^2+\dfrac{1}{x^2}\right)^2=x^2+8x+16\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(8.x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\right]+4\left(x^4+2+\dfrac{1}{x^2}\right)-x^2-8x-16=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(\dfrac{8x^2+1}{x}-4x^2-\dfrac{4}{x^2}\right)\right]+4x^4+8+\dfrac{4}{x^2}-x^2-8x-16=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{x\left(8x^2+1\right)}{x^2}-\dfrac{4x^2.x^2}{x^2}-\dfrac{4}{x^2}\right)+......=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{8x^3+x-4x^4-4}{x^2}\right)+...=0\)
\(\Leftrightarrow\dfrac{x^2}{x}.-\dfrac{4x^4+8x^3+x-4}{x^2}+.....=0\)
\(\Leftrightarrow-\dfrac{4x^6+8x^5+x^3-4x^2}{x^3}+\dfrac{4x^4+8+4x^2}{1}-\dfrac{x^2-8x-16}{1}=0\)
\(\Leftrightarrow......+\dfrac{x^3.\left(4x^4+8+4x^2\right)}{x^3}-\dfrac{x^3\left(x^2-8x-16\right)}{x^3}=0\)
\(\Leftrightarrow-4x^6+8x^5+x^3-4x^2+4x^7+8x^3+4x^5-x^5+8x^4+16x^3=0\)
\(\Leftrightarrow4x^7-4x^6+12x^5+8x^4+25x^3-4x^2=0\)
=> x=0 ( loại , ko tm)
Vậy pt vô nghiệm
2 tháng 10 2021
a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)
\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)
\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)
\(=-41x^2+34x-7-2x^3+4x-2\)
\(=-2x^3-41x^2+38x-9\)
b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)
\(=\left(3a+1+3a-1\right)^2\)
\(=36a^2\)
Q
23 tháng 6 2017
a) \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)\cdot\left[x\cdot\left(x-1\right)-\left(x^2-x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2-x-x^2+x-1\right)\)
\(=\left(x+1\right)\cdot\left(-1\right)\)
\(=-1\left(x+1\right)\)
b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x+4\right)\left(x-4\right)\)
\(=x^3-3x^2+3x-1-\left(x^3+8\right)+\left(3x+12\right)\left(x-1\right)\)
\(=x^3-3x^2+3x-1-\left(x^3+8\right)+3x^2-3x+12x-12\)
\(=x^3-1-x^3-8+12x-12\)
\(=-21+12x\)
c) \(3x^2\left(x+1\right)\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right)\left(x^4+x^2+1\right)\)
\(=3x^2\left(x^2-1\right)+x^6-3x^4+3x^2-1-\left(x^6-1\right)\)
\(=3x^4-3x^2+x^6-3x^4+3x^2-1-x^6+1\)
\(=0\)
20 tháng 8 2021
a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)
\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)
\(=8\left(7x+4\right)\)
=56x+32
b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)
\(=8x^2-32x+32-3x^2+12x+15-5x^2\)
\(=-20x+47\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)
\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)
=2
25 tháng 1 2017
áp dụng tính chất kết hợp của phép cộng các phân thức, tính dần từ trái sang phải:
\(A=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(A=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(A=\frac{32}{1-x^{32}}\)
PA
25 tháng 1 2017
\(A=\frac{x^4-\left(x-1\right)^2}{\left(x^2+1\right)^2-x^2}+\frac{x^2-\left(x^2-1\right)^2}{x^2\left(x+1\right)^2-1}+\frac{x^2\left(x-1\right)^2-1}{x^4-\left(x+1\right)^2}\)
\(=\frac{\left(x^2-x+1\right)\left(x^2+x-1\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}+\frac{\left(x-x^2+1\right)\left(x+x^2-1\right)}{\left(x^2+x-1\right)\left(x^2+x+1\right)}+\frac{\left(x^2-x-1\right)\left(x^2-x+1\right)}{\left(x^2-x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x^2+x-1}{x^2+x+1}+\frac{x-x^2+1}{x^2+x+1}+\frac{x^2-x+1}{x^2+x+1}\)
\(=\frac{x^2+x-1+x-x^2+1+x^2-x+1}{x^2+x+1}\)
\(=\frac{x^2+x+1}{x^2+x+1}\)
= 1
a.\(2^{x+1}=1\)
\(2^{x+1}=2^0\)
\(\Rightarrow x+1=0\Leftrightarrow x=-1\)
Vậy...
b.\(\left(-2\right)^{x+2}=4\)
\(\left(-2\right)^{x+2}=\left(-2\right)^2\)
\(\Rightarrow x+2=2\Leftrightarrow x=0\)
Vậy.....
\(a,2^{x+1}=1\)
\(2^{x+1}=2^0\)
\(\Rightarrow x+1=0\)
\(x=0-1\)
\(x=-1\)
\(b,\left(-2\right)^{x+2}=4\)
\(\left(-2\right)^{x+2}=\left(-2\right)^2\)
\(\Rightarrow x+2=2\)
\(x=2-2\)
\(x=0\)