Tìm GTNN
A=x^2+14x+50
B=x^2+4x+6
C=x^2-x+1
D=x^2+3x+3
E=2x^2+4x+3
K
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4 tháng 9 2021
\(A=x^2-4x+1\)
\(A=x^2-4x+4-3\)
\(A=\left(x-2\right)^2-3\)
Min A = -3
Min A xảy ra khi (x-2)2=0
x-2=0
x=2
4 tháng 9 2021
A đến C là tìm GTNN
\(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)
Dấu "=" xảy ra ⇔ x=2
\(B=2x^2-x+1=2\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)+\dfrac{7}{8}=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{4}\)
\(C=x^2-x+1=\left(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)
2 tháng 10 2021
a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)
\(\Leftrightarrow-62x=-92\)
hay \(x=\dfrac{46}{31}\)
b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)
\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)
\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)
\(\Leftrightarrow8x^2+49x-15=0\)
\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)
NT
3
27 tháng 6 2021
\(a,=3x-9-4x+12=-x+3=0\)
\(\Leftrightarrow x=3\)
Vậy ..
\(b,=\left(x+2\right)\left(x+2-x+2\right)=4\left(x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy ..
\(c,=x^3-3x^2+3x-1=\left(x-1\right)^3=0\)
\(\Leftrightarrow x=1\)
Vậy ..
\(d,\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy ..
\(e,=\left(2x-3-5\right)\left(2x-3+5\right)=\left(2x-8\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)
Vậy ...
27 tháng 6 2021
a) Ta có: 3(x-3)-4x+12=0
\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
hay x=3
Vậy: S={3}
b) Ta có: \(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow x^2+4x+4-x^2+4=0\)
\(\Leftrightarrow4x=-8\)
hay x=-2
Vậy: S={-2}
c) Ta có: \(x^3+3x=3x^2+1\)
\(\Leftrightarrow x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
Vậy: S={1}
d) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: S={0;2;-2}
T
24 tháng 10 2023
Bài 1.
a)
\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)
b)
\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)
Bài 2.
a)
\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)
b)
\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)
PJ
7 tháng 7 2019
a) 4x2 - 2x + 3 - 4x.(x - 5) = 7x - 3
--> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
--> 4x2 - 2x - 4x2 + 20x - 7x = -3 - 3
--> 11x = -6
--> x = \(\frac{-6}{11}\)
b) -3x.(x - 5) + 5.(x - 1) + 3x2 = 4x
--> -3x2 + 15x + 5x - 5 + 3x2 = 4x
--> -3x2 + 15x + 5x + 3x2 - 4x = 5
--> 16x = 5
--> x = \(\frac{5}{16}\)
c) 7x.(x - 2) - 5.(x - 1) = 21x2 - 14x2 + 3
--> 7x2 - 14x - 5x + 5 = 7x2 + 3
--> 7x2 - 14x - 5x - 7x2 = -5 + 3
--> -19x = -2
--> x = \(\frac{2}{19}\)
d) 3.(5x - 1) - x.(x - 2) + x2 - 13x = 7
--> 15x - 3 - x2 + 2x + x2 - 13x = 7
--> 15x - x2 + 2x + x2 - 13x = 3 + 7
--> 4x = 10
--> x = \(\frac{5}{2}\)
e) \(\frac{1}{5}\)x.(10x - 15) - 2x.(x - 5) = 12
--> 2x2 - 3x - 2x2 + 10x = 12
--> 7x = 12
--> x = \(\frac{12}{7}\)
~ Học tốt ~
4 tháng 7 2019
a) 4x2 - 2x + 3 - 4x(x - 5) = 7x - 3
=> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
=> 18x + 3 = 7x - 3
=> 18x - 7x = -3 - 3
=> 11x = -6
=> x = -6/11
b) -3x(x - 5) + 5(x - 1) + 3x2 = 4x
=> -3x2 + 15x + 5x - 5 + 3x2 = 4x
=> 20x - 5 = 4x
=> 20x - 4x = 5
=> 16x = 5
=> x = 5/16
4 tháng 7 2019
\(c,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)
\(\Leftrightarrow7x^2-14x-5x+5=7x^2+3\)
\(\Leftrightarrow7x^2-7x^2-19x=3-5\)
\(\Leftrightarrow-19x=-2\)
\(\Leftrightarrow x=\frac{2}{19}\)
A=(x+7)^2+1>=1
Dấu = xảy ra khi x=-7
B=(x+2)^2+2>=2
Dấu = xảy ra khi x=-2
C=(x-1/2)^2+3/4>=3/4
Dấu = xảy ra khi x=1/2
D=(x+3/2)^2+3/4>=3/4
Dấu = xảy ra khi x=-3/2
E=2(x+1)^2+1>=1
Dấu = xảy ra khi x=-1
A = x2 + 14x + 50 = (x2 + 14x + 49) + 1 = (x + 7)2 + 1
Ta có: (x + 7)2 \(\ge\)0 \(\forall\)x
=> (x + 7)2 + 1 \(\ge\)1 \(\forall\)x
Dấu "=" xảy ra <=> x + 7 = 0 <=> x = -7
Vậy MinA = 1 <=> x = -7
B = x2 + 4x + 6 = (x2 + 4x + 4) + 2 = (x + 2)2 + 2
Ta luôn có : (x + 2)2 \(\ge\)0 \(\forall\)x
=> (x + 2)2 + 2 \(\ge\)2 \(\forall\)x
Dấu "=" xảy ra <=> x + 2 = 0 <=> x = -2
Vậy MinB = 2 <=> x = -2
C = x2 - x + 1 = (x2 - x + 1/4) + 3/4 = (x - 1/2)2 + 3/4
Ta luôn có: (x - 1/2)2 \(\ge\)0 \(\forall\)x
=> (x - 1/2)2 + 3/4 \(\ge\)3/4 \(\forall\)x
Dấu "=" xảy ra <=> x - 1/2 = 0 <=> x = 1/2
Vậy MinC = 3/4 <=> x = 1/2