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bài 1:
a) 4n+4+3n-6<19
<=> 7n-2<19
<=> 7n<21 <=> n< 3
b) n\(^2\) - 6n + 9 - n\(^2\) + 16\(\leq\)43
-6n+25\(\leq\)43
-6n\(\leq\)18
n\(\geq\)-3
Câu 2:
\(A=3\left(2x+9\right)^2-1>=-1\)
Dấu '=' xảy ra khi x=-9/2
Câu 9:
=>(x-30)^2=0
=>x-30=0
=>x=30
Câu 10:
\(=2x^2+6x-4x-12-2x^2-2x=-12\)
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
a, \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
\(=\left(1-\dfrac{1}{4}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
\(=\left(1-\dfrac{1}{16}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
...
\(=\left(1-\dfrac{1}{2^{2n}}\right)\left(1+\dfrac{1}{2^{2n}}\right).2=\left(1-\dfrac{1}{2^{4n}}\right).2=2-\dfrac{1}{2^{4n-1}}\)
Vậy ...
b,Sửa đề: \(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)
Ta có:\(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)
\(=\left(10-1\right).\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
\(=\left(10^2-1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
\(=\left(10^4-1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
...
\(=\left(10^{2n}-1\right)\left(10^{2n}+1\right).\dfrac{1}{9}=\left(10^{4n}-1\right).\dfrac{1}{9}=\dfrac{10^{4n}}{9}-\dfrac{1}{9}\)
Vậy ...
áp dụng hằng đẳng thức (a+b)(a-b)=a^2-b^2 Minh Hoang Hai
Câu 1:
a: \(A=\dfrac{x+1-x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x^2+1-2x}{2}\)
\(=\dfrac{2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}=\dfrac{x-1}{x+1}\)
b: Để A=x/6 thì \(\dfrac{x-1}{x+1}=\dfrac{x}{6}\)
\(\Leftrightarrow x^2+x-6x+6=0\)
=>x=3 hoặc x=2
1)
\(M=\dfrac{1}{3}x^2+2x+10\)
\(=\dfrac{1}{3}.\left(x^2+6x+30\right)\)
\(=\dfrac{1}{3}\left(x^2+2.x.3+9\right)+7\)
\(=\dfrac{1}{3}.\left(x+3\right)^2+7\) \(\ge\) 7 với \(\forall\) x
=> M luôn dương
=> đpcm
2)
a) \(2x-x^2-15\)
\(=-\left(x^2-2x+15\right)\)
\(=-\left(x^2-2x+1\right)-14\)
\(=-\left(x-1\right)^2-14\) \(\le-14\) với \(\forall\) x
=> \(2x-x^2-15\) luôn âm
=> đpcm
b) \(-5-\left(x-1\right)\left(x+2\right)\)
\(=-5-x^2-2x+x+2\)
\(=-x^2-x-3\)
\(=-\left(x^2+x+3\right)\)
\(=-\left(x^2+2.\dfrac{1}{2}.x+\dfrac{1}{4}\right)-\dfrac{11}{4}\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\) với \(\forall\) x
=> \(-5-\left(x-1\right)\left(x+2\right)\) luôn âm
=> đpcm
\(M=\dfrac{1}{3}x^2+2x+10=\dfrac{1}{3}\left(x^2+6x+9\right)+7\)
\(=\dfrac{1}{3}\left(x+3\right)^2+7\)
Ta có:
\(\dfrac{1}{3}\left(x+3\right)^2\ge\forall x\Rightarrow\dfrac{1}{3}\left(x+3\right)^2+7>0\)
=>đpcm
\(2,a,2x-x^2-15\)
\(=-\left(x^2-2x+1\right)-14\)
\(=-\left(x-1\right)^2-14\)
Ta có:
\(-\left(x-1\right)^2\le0\forall x\Rightarrow-\left(x-1\right)^2-14< 0\)
=> đpcm
\(b,-5-\left(x-1\right)\left(x+2\right)\)
\(=-5-\left(x^2+x-2\right)\)
\(=-5-x^2-x+2\)
\(=-\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{11}{4}\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{11}{4}\)
Ta có:
\(-\left(x+\dfrac{1}{2}\right)^2\le0\forall x\Rightarrow-\left(x+\dfrac{1}{2}\right)-\dfrac{11}{4}< 0\)=> đpcm
`(2m-10)^2/((m+5)^2+1)`
`=(2m-10)^2/(m^2+10m+26)-404+404`
`=(4m^2-40m+100)/(m^2+10m+26)-404+404`
`=(4m^2-40m+100-404m^2-4040m-10504)/(404[(m+5)^2+1])+404`
`=(-400m^2-4080m-10404)/(404[(m+5)^2+1])+404`
`=(-(400m^2+4080m+10404))/(404[(m+5)^2+1])+404`
`=(-(20m+102)^2)/(404[(m+5)^2+1])+404<=404`
Dấu "=" xảy ra khi `20m+102=0<=>m=(-51)/10`
Bài này giải kiểu lớp 8 thì nó cực kì vô duyên:
\(P=\dfrac{4m^2-40m+100}{m^2+10m+26}=\dfrac{404\left(m^2+10m+26\right)-4\left(100m^2+1020m+2601\right)}{m^2+10m+26}\)
\(P=404-\dfrac{4\left(10m+51\right)^2}{\left(m+5\right)^2+1}\le404\)
\(P_{max}=404\) khi \(m=-\dfrac{51}{10}\)