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(3x-1)^2 - 16 = (3x-1)^2 - 4^2
= (3x-1-4)(3x-1+4)
= (3x-5)(3x+3)
\(\left(3x-1\right)^2-16\)
\(=\left(3x-1\right)^2-4^2\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x-5\right)\left(3x+3\right)\)
\(=3\left(x+1\right)\left(3x-5\right)\)
`\sqrt{17-12\sqrt{2}}`
`=\sqrt{17-6.\sqrt{4}.\sqrt{2}}`
`=\sqrt{17-6\sqrt{8}}`
`=\sqrt{9-2.3.\sqrt{8}+8}`
`=\sqrt{(3-\sqrt{8})^{2}}`
`=|3-\sqrt{8}|`
`=3-\sqrt{8}` ( Vì `3=\sqrt{9}>\sqrt{8}` )
Lời giải:
a. $\frac{3}{-2}=\frac{9}{-6}$
$\frac{-2}{3}=\frac{-6}{9}$
$\frac{3}{9}=\frac{-2}{-6}$
$\frac{9}{3}=\frac{-6}{-2}$
b.
$\frac{12}{16}=\frac{3}{4}$
$\frac{16}{12}=\frac{4}{3}$
$\frac{12}{3}=\frac{16}{4}$
$\frac{3}{12}=\frac{4}{16}$
c) \(\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\)
c) \(\left(x^2+3^2\right)^2=x^4+18x+81\)
c) \(\left(2x^2+1\right)^2=4x^4+4x^2+1\)
c) \(\left(3x-y^2\right)^2=9x^2-6xy^2+y^4\)
c) \(\left(x+2y^2\right)^2=x^2+4xy^2+4y^4\)
c) \(\left(3x\right)^2-y^2=\left(3x-y\right)\left(3x+y\right)\)
c) \(\left(2x+3y^2\right)^2=4x^2+12xy^2+9y^4\)
c) \(\left(4x-2y^2\right)^2=16x^2-16xy^2+4y^4\)
c) \(\left(4x^2-2y\right)^2=16x^4-16x^2y+4y^2\)
c) \(\left(\dfrac{1}{x}-5\right)\left(\dfrac{1}{x}+5\right)=\dfrac{1}{x^2}-25\)
c) \(\left(x-\dfrac{3}{2}\right)\left(x+\dfrac{3}{2}\right)=x^2-\dfrac{9}{4}\)
c) \(\left(\dfrac{x}{3}-\dfrac{y}{4}\right)\left(\dfrac{x}{3}+\dfrac{y}{4}\right)=\dfrac{x^2}{9}-\dfrac{y^2}{16}\)
c) \(\left(\dfrac{x}{y}-\dfrac{2}{3}\right)\left(\dfrac{x}{y}+\dfrac{2}{3}\right)=\dfrac{x^2}{y^2}-\dfrac{4}{9}\)
c) \(\left(\dfrac{x}{2}+\dfrac{y}{3}\right)\left(\dfrac{y}{3}-\dfrac{x}{2}\right)=\dfrac{y^2}{9}-\dfrac{x^2}{4}\)
c) \(\left(2x-\dfrac{2}{3}\right)\left(\dfrac{2}{3}+2x\right)=4x^2-\dfrac{4}{9}\)
c) \(\left(2x+\dfrac{3}{5}\right)\left(\dfrac{3}{5}-2x\right)=\dfrac{9}{25}-4x^2\)
c) \(\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)\left(\dfrac{4}{4}+\dfrac{1}{2}x\right)=\dfrac{1}{4}x^2-\dfrac{16}{9}\)
c) \(\left(\dfrac{2}{3}x^2-\dfrac{y}{2}\right)\left(\dfrac{2}{3}x^2+\dfrac{y}{2}\right)=\dfrac{4}{9}x^4-\dfrac{y^2}{4}\)
\(299.301=\left(300-1\right)\left(300+1\right)=300^2-1^2=90000-1=89999\)
\(a\text{) }pt\Leftrightarrow\left(y^2+2y+1\right)+\left[\left(2^x\right)^2-2.2^x+1\right]=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
\(\Leftrightarrow y+1=0\text{ và }2^x-1=0\)
\(\Leftrightarrow y=-1\text{ và }x=0\)
\(b\text{) }pt\Leftrightarrow\left(4x^2+4y^2+8xy\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow x+y=0\text{ và }x-1=0\text{ và }y+1=0\)
\(\Leftrightarrow x=1\text{ và }y=-1\)
\(-4.20=-16.5\)
\(\Rightarrow\frac{-4}{-5}=\frac{-16}{20};\frac{-5}{-4}=\frac{20}{-16};\frac{-16}{-4}=\frac{20}{-5};\frac{-4}{-16}=\frac{-5}{20}\)
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