(x+8)^2=4x^2-4x+1
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1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
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a) x² + 6x + 8
= x² + 2x + 4x + 8
= (x² + 2x) + (4x + 8)
= x(x + 2) + 4(x + 8)
= (x + 2)(x + 4)
b) 3x² - 2(x - y)² - 3y²
= (3x² - 3y²) - 2(x - y)²
= 3(x² - y²) - 2(x - y)²
= 3(x + y)(x - y) - 2(x - y)²
= (x - y)[3(x + y) - 2(x - y)]
= (x - y)(3x + 3y - 2x + 2y)
= (x - y)(x + 5y)
c) 4x² - 9y² + 4x - 6y
= (4x² - 9y²) + (4x - 6y)
= (2x - 3y)(2x + 3y) + 2(2x - 3y)
= (2x - 3y)(2x + 3y + 2)
d) x(x + 1)² + x(x - 5) - 5(x + 1)²
= [x(x + 1)² - 5(x + 1)²] + x(x - 5)
= (x + 1)²(x - 5) + x(x - 5)
= (x - 5)[(x + 1)² + x]
= (x - 5)(x² + 2x + 1 + x)
= (x - 5)(x² + 3x + 1)
e) 2xy - x² + 3y² - 4y + 1
= -x² + 2xy - y² + 4y² - 4y + 1
= -(x² - 2xy + y²) + (4y² - 4y + 1)
= -(x - y)² + (2y - 1)²
= (2y - 1)² - (x - y)²
= (2y - 1 - x + y)(2y - 1 + x - y)
= (3y - x - 1)(x + y - 1)
f) 4x¹⁶ + 81
= (2x⁸)² + 2.2x⁸.9 + 9² - 2.2x⁸.9
= (2x⁸ + 9)² - 36x⁸
= (2x⁸ + 9) - (6x⁴)²
= (2x⁸ + 9 - 6x⁴)(2x⁸ + 9 + 6x⁴)
= (2x⁸ - 6x⁴ + 9)(2x⁸ + 6x⁴ + 9)
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a, \(x^2+4x-5=x^2+2x+2x+4-9\)
\(=\left(x^2+2x\right)+\left(2x+4\right)-9\)
\(=x.\left(x+2\right)+2.\left(x+2\right)-9\)
\(=\left(x+2\right)^2-9\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2-9\ge-9\) với mọi giá trị của \(x\in R\).
Để \(\left(x+2\right)^2-9=-9\) thì \(\left(x+2\right)^2=0\Rightarrow x=-2\)
Vậy.......
b, \(4x^2+4x-3=4x^2+2x+2x+1-4\)
\(=2x.\left(2x+1\right)+\left(2x+1\right)-4\)
\(=\left(2x+1\right)^2-4\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2-4\ge-4\) với mọi giá trị của \(x\in R\).
Để \(\left(2x+1\right)^2-4=-4\) thì \(\left(2x+1\right)^2=0\Rightarrow x=\dfrac{-1}{2}\)
Vậy.........
c, \(x^2+x+1=x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=x.\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}.\left(x+\dfrac{1}{2}\right)+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x+\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) với mọi giá trị của \(x\in R\).
Để \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\) thì \(\left(x+\dfrac{1}{2}\right)^2=0\Rightarrow x=\dfrac{-1}{2}\)
Vậy.........
Chúc bạn học tốt!!!
Các câu còn lại làm tương tự!!
a) A = x2 + 4x - 5
A = x2 + 4x + 4 +1 = ( x + 2 )2 + 1 \(\ge\) 1 với mọi x
MinA = 1 khi và chỉ khi x = -2
b) B = 4x2 + 4x - 3
B = 4x2 + 4x + 1 - 4
B = ( 2x+1 )2 - 4 \(\ge\) -4 với mọi x
MinB = -4 khi và chỉ khi x = \(\dfrac{-1}{2}\)
c) C = x2 + x + 1
C = x2 + x + \(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)
C = ( x + \(\dfrac{1}{2}\) )2 + \(\dfrac{3}{4}\) \(\ge\) \(\dfrac{3}{4}\) với mọi x
MinC = \(\dfrac{3}{4}\) khi và chỉ khi x = \(-\dfrac{1}{2}\)
d) D = 2x2 + 4x + 8
D = 2 . ( x2 + 2x + 4 )
D = 2. ( x2 + 2x + 1 + 3 )
D = 2. \(\left[\left(x+1\right)^2+3\right]\)
D = 2.( x+1 )2 + 6 \(\ge\) 6 với mọi x
MinD = 6 khi và chỉ khi x = -1
e) E = x2 + x
E = x2 + x + \(\dfrac{1}{4}\) - \(\dfrac{1}{4}\)
E = \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\) \(\ge\) \(-\dfrac{1}{4}\) với mọi x
MinE = \(-\dfrac{1}{4}\) khi và chỉ khi x = \(\dfrac{-1}{2}\)
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a: \(\dfrac{x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)
\(=\dfrac{x+10}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-\left(x+10\right)}{2\left(x+2\right)}\)
b: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)
\(=\dfrac{\left(2x-1\right)\left(2x+1\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(x-2\right)}\)
\(=\dfrac{3\left(2x-1\right)\left(2x+1\right)}{2\left(x-2\right)\left(x+4\right)}\)
c: \(=\dfrac{4y^2}{7x^4}\cdot\dfrac{35x^2}{-8y}=\dfrac{5}{x^2}\cdot\dfrac{-1}{2}\cdot y=\dfrac{-5y}{2x^2}\)
d: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)
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ta có
\(5x=-3y=4z\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{3z}{45}=\frac{x-y+3z}{12+20+45}=\frac{7}{77}=\frac{1}{11}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{11}.12=\frac{12}{11}\\-y=\frac{1}{11}.20=\frac{20}{11}\\3z=\frac{1}{11}.45=\frac{45}{11}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{12}{11}\\y=-\frac{20}{11}\\z=\frac{45}{11}:3=\frac{15}{11}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=\frac{12}{11}\\y=\frac{-20}{11}\\z=\frac{15}{11}\end{cases}}\)
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a.(x+10) /(4*x)-8* 4 -(2*x)/x+2
-(127*x-10)/(4*x)
(5/2-127*x/4)/x
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c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)