(16a⁴+b⁴+4a²b²):(4a²+b²-2ab)
Những câu hỏi liên quan
cho 4a^2 +b^2=5ab với 4a>b>0. tính gia trị bt: Q= 5ab/(16a^2 - b^2)
Ta có: 4a2+b2=5ab
=>(4a2-5ab+b2)=0
=>(4a2-4ab)-(ab-b2)=0
=>4a(a-b)-b(a-b)=0
=>(4a-b)(a-b)=0
=>4a=b hoặc a=b
Mà 4a>b
=>a=b
=>\(\frac{5ab}{16a^2-b^2}=\frac{5a^2}{16a^2-a^2}=\frac{5a^2}{15a^2}=\frac{1}{3}\)
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cho 4a^2 +b^2 = 5ab với 4a>b>0. tính gia trị bt: Q= 5ab/(16a^2 -b^2)
\(\sqrt{4x^2-4x+1}+2=3x\)
Rút gọn
\(7\sqrt{a}-5b\sqrt{16a^3}+4a\sqrt{25ab^2}-3\sqrt{16a}\) với a>0, b>0
\(\sqrt{4x^2-4x+1}+2=3x\)
Vì \(VT\ge2\Rightarrow VP\ge2\Rightarrow x\ge\dfrac{2}{3}\)
\(\Rightarrow\sqrt{\left(2x-1\right)^2}+2=3x\Rightarrow\left|2x-1\right|+2=3x\)
\(\Rightarrow2x-1+2=3x\left(x\ge\dfrac{2}{3}\right)\Rightarrow x=1\)
\(7\sqrt{a}-5b\sqrt{16a^3}+4a\sqrt{25ab^2}-3\sqrt{16a}\)
\(=7\sqrt{a}-20ab\sqrt{a}+20ab\sqrt{a}-12\sqrt{a}=-5\sqrt{a}\)
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cho 2 số a,b thỏa: a2-2ab+1=2(ab-b2)
tinhsP= \(\dfrac{a^5+b^5+2ab}{4a^3-2ab}\)
C/m rằng
a) (a+3) .( a^2 -3a+9) - (54+x^3)
b) (2a+b) . (4a^2 -2ab +b^2 ) - (2a-b) . (4a^2+ +2ab +b^2)
c) (x +y)^2 - (x-y)^2
d) (x+y)^3 -( x-y)^3 -2y^3
[TEX]frac{a}{c} frac{b}{d}[/TEX]
Rightarrow [TEX]adcb[/TEX]
Rightarrow [TEX]2ad2cb[/TEX]
Rightarrow [TEX]ad-cbcb-ad[/TEX]
Rightarrow[TEX]ab(ad-cb)ab(cb-ad)[/TEX]
Rightarrow[TEX]a^2bd- acb^2 acb^2-a^2bd[/TEX]
Rightarrow[TEX]16a^2b^2-36abcd+24a^2bd-24acb^2 16a^2b^2- 36abcd+24acb^2-24a^2bd[/TEX]
Rightarrow[TEX]( 4a^2 - 6ac)(4b^2 + 6bd) ( 4a^2 +6ac)(4b^2 - 6bd)[/TEX]
C2: [TEX]frac{a}{c} frac{b}{d}[/TEX]Rightarrow[TEX] (frac{a}{b})^2( frac{c}{d})^2 frac{ac}{bd}frac{-12ac}{-12bd} frac{(4a^2-6ac)-...
Đọc tiếp
\( [TEX]\frac{a}{c} = \frac{b}{d}[/TEX] \Rightarrow [TEX]ad=cb[/TEX] \Rightarrow [TEX]2ad=2cb[/TEX] \Rightarrow [TEX]ad-cb=cb-ad[/TEX] \Rightarrow[TEX]ab(ad-cb)=ab(cb-ad)[/TEX] \Rightarrow[TEX]a^2bd- acb^2= acb^2-a^2bd[/TEX] \Rightarrow[TEX]16a^2b^2-36abcd+24a^2bd-24acb^2 = 16a^2b^2- 36abcd+24acb^2-24a^2bd[/TEX] \Rightarrow[TEX]( 4a^2 - 6ac)(4b^2 + 6bd) =( 4a^2 +6ac)(4b^2 - 6bd)[/TEX] C2: [TEX]\frac{a}{c} = \frac{b}{d}[/TEX]\Rightarrow[TEX] (\frac{a}{b})^2=( \frac{c}{d})^2= \frac{ac}{bd}=\frac{-12ac}{-12bd}= \frac{(4a^2-6ac)-(4a^2+6ac)}{(4b^2- 6bd)- (4b^2+6bd)} [/TEX]\)
Tìm a,b thuộc Z biết:
a)2ab-b+3a=10
b)a2-2ab+4a-8b=7
Cho a,b,c >0 và \(\frac{b-20a+16c}{4a}=\frac{c-20b+16a}{4b}=\frac{a-20c+16b}{4c}\)
Tính giá trị \(F=\left(4+\frac{a}{4b}\right).\left(4+\frac{b}{4c}\right).\left(4+\frac{c}{4a}\right)\)
Trừ mỗi vế cho 1, ta có:
\(\frac{b-16a+16c}{4a}=\frac{c-16b+16a}{4b}=\frac{a-16c+16b}{4c}=\frac{a+b+c}{4.\left(a+b+c\right)}=\frac{1}{4}\)(vì a,b,c > 0 nên a+b+c>0)
\(\Leftrightarrow\hept{\begin{cases}b+16c=17a\\c+16a=17b\\a+16b=17c\end{cases}}\Leftrightarrow a=b=c\)
tự thay vào
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√ 4a + √ 16a (a>=0)
\(\sqrt{4a}+\sqrt{16a}=\sqrt{2^2a}+\sqrt{4^2a}=2a+4a=6a\)
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