(x-1)^x+2=(x-1)^x+6
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1x2= 2 1x2x3=6 1x2x3x4=24 1x2x3x4x5=120 1x2x3x4x5x6=720 1x2x3x4x5x6x7=5040
1x2x3x4x5x6x7x8=40320 1x2x3x4x5x6x7x8x9=362880 1x2x3x4x5x6x7x8x9x10=3628800
1 x 2 = 2
1 x 2 x 3 = 6
1 x 2 x 3 x 4 = 24
1 x 2 x 3 x 4 x 5 = 120
1 x 2 x 3 x 4 x 5 x 6 = 720
1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 362880
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3628800
![](https://rs.olm.vn/images/avt/0.png?1311)
x2-4x+7 = 0 ⇔ x2 -4x + 4 + 3 = 0
⇔ (x-2)2+3=0 ⇔ (x-2)2=-3 (vô lí)
Vậy pt vô nghiệm
*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm
Ta có: \(x^2-4x+7=0\)
\(\Leftrightarrow x^2-4x+4+3=0\)
\(\Leftrightarrow\left(x-2\right)^2+3=0\)
mà \(\left(x-2\right)^2+3\ge3>0\forall x\)
nên \(x\in\varnothing\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
MẤY DÒNG NÀO BẠN THẤY KO CẦN THIẾT THÌ LƯỢC BỎ NHA!!!
a) \(2\left(x-5\right)-3\left(x+6\right)=4\left(x-7\right)\)
\(2x-10-3x-18=4x-28\)
\(2x-3x-4x-10-18=-28\)
\(-5x-28=-28\)
\(-5x=-28+28=0\)
\(x=\frac{0}{-5}=0\)
b) \(3\left(x-1\right)-2\left(x+5\right)=2\left(x-3\right)\)
\(3x-3-2x-10=2x-6\)
\(3x-2x-2x-3-10=-6\)
\(-x-13=-6\)
\(-x=-6+13=7\)
\(x=-7\)
c) \(5\left(1-x\right)-6\left(1+x\right)=7\left(3-x\right)\)
\(5-5x-6-6x=21-7x\)
\(-5x-6x+7x+5-6=21\)
\(-4x-1=21\)
\(-4x=22\)
\(x=\frac{22}{-4}=\frac{-11}{2}\)
d) \(2x+5-3\left(3x+7\right)=6\left(1-x\right)+8\)
\(2x+5-9x-21=6-6x+8\)
\(2x-9x+6x+5-21=6+8\)
\(-x-16=14\)
\(-x=14+16=30\)
\(x=-30\)
e) \(x-2+3\left(x-4\right)=5\left(x-6\right)+7\)
\(x-2+3x-12=5x-30+7\)
\(x+3x-5x-2-12=-30+7\)
\(-x-14=-23\)
\(-x=-23+14=-9\)
\(x=9\)
f) \(x+2+3\left(1-x\right)-5\left(2-x\right)=6\left(1-x\right)+\left(3-x\right)\)
\(x+2+3-3x-10+5x=6-6x+3-x\)
\(x-3x+5x+6x+x+2+3-10=6+3\)
\(10x-7=9\)
\(10x=9+7=16\)
\(x=\frac{16}{10}=\frac{8}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d: ĐKXĐ: \(x\notin\left\{2;-3\right\}\)
\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)
=>\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{-5}{\left(x+3\right)\left(x-2\right)}\)
=>\(x+3-6\left(x-2\right)=-5\)
=>x+3-6x+12=-5
=>-5x+15=-5
=>-5x=-20
=>x=4(nhận)
e: ĐKXĐ: x<>-2
\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)
=>\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{5}{x^2-2x+4}\)
=>\(2\left(x^2-2x+4\right)-2x^2-16=5\left(x+2\right)\)
=>\(2x^2-4x+8-2x^2-16=5x+10\)
=>5x+10=-4x-8
=>9x=-18
=>x=-2(loại)
f: ĐKXĐ: \(x\in\left\{1;-1\right\}\)
\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)
\(\Leftrightarrow\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
=>\(\dfrac{\left(x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
=>\(\left(x^3+1\right)\left(x^2-1\right)-\left(x^3-1\right)\left(x^2-1\right)=2\left(x^2+4x+4\right)\)
=>\(\left(x^2-1\right)\cdot\left(x^3+1-x^3+1\right)=2\left(x^2+4x+4\right)\)
=>\(2x^2+8x+8=\left(x^2-1\right)\cdot2=2x^2-2\)
=>8x=-10
=>x=-5/4(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{6x^2+8x+7}{x^3-1}+\frac{x}{x^2+x+1}+\frac{6}{1-x}\right)\left(x^2-1\right)\)
\(=\left[\frac{6x^2+8x+7}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\left(x-1\right)\left(x+1\right)\)
\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\left(x-1\right)\left(x+1\right)=x+1\)
bài này sai đề rùi