B = 5 - 4x2 + 5
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a) 125x57+27x5^4+5^2x40
=5^3x57+27x5^4+5^2x40
=5^2x(5x57+27x5^2+40)
=25x1000
=25000
\(\dfrac{3}{5}+\dfrac{5}{7}\) x \(\dfrac{3}{5}-\dfrac{9}{7}\) x \(\dfrac{3}{5}\)
\(=\dfrac{3}{5}+\left(\dfrac{5}{7}-\dfrac{9}{7}\right)\) x \(\dfrac{3}{5}\)
\(=\dfrac{3}{5}-\dfrac{4}{7}\) x \(\dfrac{3}{5}\)
\(=\dfrac{3}{5}-\dfrac{12}{35}\)
\(=\dfrac{21}{35}-\dfrac{12}{35}=\dfrac{21-12}{35}=\dfrac{9}{35}\)
a (3-2x)2 = 6 - 4x
b (xy+5)2 = 2xy + 10
c (2x+1)(1-2x) = 2x - 4x2 + 1 - 2x = 4x2 + 1
d (1-5x)3 = 3-15x
e (2x+y)(4x2 - 4xy + y2) = 8x3 -8x2y+2xy2 + 4x2y-4xy2 + y3 = 8x3 + y3 - 4x2y - 2xy2
\(a,VT=9+4\sqrt{5}=\sqrt{5^2}+2.2\sqrt{5}+2^2=\left(\sqrt{5}+2\right)^2=VP\left(dpcm\right)\)
\(b,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\Leftrightarrow\sqrt{9-4\sqrt{5}}=\sqrt{5}-2\)
Ta có : \(VT=\sqrt{9-4\sqrt{5}}=\sqrt{\sqrt{5^2}-2.2\sqrt{5}+2^2}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2=VP\left(dpcm\right)\)
a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)
\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)
b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)
\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)
c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)
\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)
d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)
\(ĐTXR\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a: Ta có: \(4x^2+12x+1\)
\(=4x^2+12x+9-8\)
\(=\left(2x+3\right)^2-8\ge-8\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)
b: Ta có: \(4x^2-3x+10\)
\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)
\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)
\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)
c: Ta có: \(2x^2+5x+10\)
\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)
\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)
\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)
Ta có: \(\left(2x-5\right)\left(4x^2+10x+25\right)\left(2x+5\right)\left(4x^2-10x+25\right)-64x^6\)
\(=\left(8x^3-125\right)\left(8x^3+125\right)-64x^6\)
\(=64x^6-15625-64x^6\)
=-15625
a) C1 : 2/5 x 3/7 + 2/7 x 4/7
= 2/5 x (3/7 + 4/7)
= 2/5 x 1
= 2/5
C2 : 2/5 x 3/7 + 2/5 x 4/7
= 6/35 + 8/35
= 2/5
b) (2/3 - 4/7) x 5/5
= 2/21 x 5/5
= 2/21
C2 : (2/3 - 4/7) x 5/5
= 2/3 x 1 - 4/7 x 1
= 2/3 - 4/7
= 2/21
c) 3/4 x 2/5 - 3/4 x 2/7
= 3/4 x (2/5 - 2/7)
= 3/4 x 4/35
= 3/35
C2 : 3/4 x 2/5 - 3/4 x 2/7
= 3/10 - 3/14
= 3/35
a,
Cách 1
\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)
\(=\frac{6}{35}+\frac{8}{35}\)
\(=\frac{2}{5}\)
Cách 2 :
\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)
\(=\frac{2}{5}\times\left(\frac{3}{7}+\frac{4}{7}\right)\)
\(=\frac{2}{5}\times1\)
\(=\frac{2}{5}\)
b,
Cách 1 :
\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)
\(=\frac{2}{21}\times1\)
\(=\frac{2}{21}\)
Cách 2 :
\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)
\(=\frac{2}{3}\times\frac{5}{5}-\frac{4}{7}\times\frac{5}{5}\)
\(=\frac{2}{3}-\frac{4}{7}\)
\(=\frac{2}{21}\)
c,
Cách 1 :
\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)
\(=\frac{3}{10}-\frac{3}{14}\)
\(=\frac{3}{35}\)
Cách 2 :
\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)
\(=\frac{3}{4}\times\left(\frac{2}{5}-\frac{2}{7}\right)\)
\(=\frac{3}{4}\times\frac{4}{35}\)
\(=\frac{3}{35}\)
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