a) x^2 + 20x + 100
b) x^2 - 20xy + 100y^2
c) x^2 - 20x + 100
Câu 9: Đa thức x2 + 20x + 100 được phân tích thành nhân tử có kết quả là
A. (x + 10) 2 B. (x + 50) 2 C. x(x + 20) + 100
Tìm x
a) x2 + 10x = 0
b) (x - 7)3 = (x - 7)
c) x2 - 20x + 100 = 0
a) \(x^2+10x=0\)
\(x\left(x+10\right)=0\)
\(\left[{}\begin{matrix}x=0\\x=-10\end{matrix}\right.\)
b) \(\left(x-7\right)^3=x-7\)
\(\left(x-7\right)^3-\left(x-7\right)=0\)
\(\left(x-7\right)\left[\left(x-7\right)^2-1\right]=0\)
\(\left(x-7\right)\left(x-7-1\right)\left(x-7+1\right)=0\)
\(\left(x-7\right)\left(x-8\right)\left(x-6\right)=0\)
\(\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
c) \(x^2-20x+100=0\)
\(x^2-10x-10x+100=0\)
\(x\left(x-10\right)-10\left(x-10\right)=0\)
\(\left(x-10\right)\left(x-10\right)=0\)
\(\left(x-10\right)^2=0\)
=> x = 10
a) \(x^2+10x=0\)
\(\Leftrightarrow x\left(x+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+10=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-10\end{matrix}\right.\)
Vậy..
b) \(\left(x-7\right)^3=\left(x-7\right)\)
\(\Leftrightarrow\left(x-7\right)^3-\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left[\left(x-7\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-8\right)\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-7=0\\x-8=0\\x-6=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
Vậy..
c) \(x^2-20x+100=0\)
\(\Leftrightarrow\left(x-10\right)^2=0\)
\(\Rightarrow x-10=0\)
\(\Rightarrow x=10\)
a, \(x^2+10x=0\)
\(\Leftrightarrow x\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-10\end{matrix}\right.\)
Vậy x = 0 hoặc x = -10
b, \(\left(x-7\right)^3=\left(x-7\right)\)
\(\Leftrightarrow\left(x-7\right)^3-\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left[\left(x-7\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\\left(x-7\right)^2-1=0\end{matrix}\right.\)
+) \(x-7=0\Leftrightarrow x=7\)
+) \(\left(x-7\right)^2-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=6\end{matrix}\right.\)
Vậy x = 7 hoặc x = 8 hoặc x = 6
c, \(x^2-20x+100=0\)
\(\Leftrightarrow\left(x-10\right)^2=0\)
\(\Leftrightarrow x=10\)
Vậy x = 10
Bài 2
a) A=x^2 - 20x + 100 tại x bằng 20
b) B=(x+2)^2 +2 (x+2) (3x + 5) + (3x+5)+(3x+5)^2 tại x=-1/4
A = x2 - 20x + 100 = x2 - 2.x.10 + 102 = ( x - 10 )2 = ( 20 - 10 )2 = 102 = 100
B = ( x + 2 )2 + 2( x + 2 )( 3x + 5 ) + ( 3x + 5 )2 ( vầy thôi nhỉ ? )
= [ ( x + 2 ) + ( 3x + 5 ) ]2
= ( x + 2 + 3x + 5 )2
= ( 4x + 7 )2
= \(\left[4\cdot\left(-\frac{1}{4}\right)+7\right]^2\)
= ( -1 + 7 )2 = 62 = 36
Tìm GTLN B=\(\frac{x}{x^2+20x+100}\)
Để \(B\)lớn nhất thì \(\frac{1}{B}\) nhỏ nhất
Ta có: \(\frac{1}{B}=\frac{x^2+20x+100}{x}=x+\frac{100}{x}+20\)
Áp dụng BĐT Cô-si ta có: \(\frac{1}{B}=x+\frac{100}{x}+20\ge2\sqrt{x.\frac{100}{x}}+20=2.\sqrt{100}+20=40\)
Dấu :'=" xảy ra \(\Leftrightarrow\)\(x=\frac{100}{x}\)\(\Leftrightarrow\)\(x=10\)
Min \(\frac{1}{B}=40\) \(\Rightarrow\) Max \(B=\frac{1}{40}\) \(\Leftrightarrow\)\(x=10\)
P/s: tham khảo nhé, nếu có sai đâu m.n chỉ mk nhé (yếu nhất cực trị)
chỉ giúp mình . ghi rõ nha
a). x+2x+3x+......+20x=2100
b). x+( x+1)+(x+2)+....+(x+100)=100x+600
c). (x-3) * (x-2)=0
d). x*2x-6x=0
Tìm giá trị nhỏ nhất , lớn nhất : C= 16x2 - 8x + 2024
D= -25x2 + 50x - 2023
B=-x2 + 20x + 100
E=(2x - 1 )2 - ( 3 x + 2 ) nhân ( x - 5 )
F=( 3 x - 5 ) 2 - ( 3x + 2 ) nhân ( 4x - 1)
mk đang cần gấp mn giúp mình vs ạ
\(C=16x^2-8x+2024\)
\(\Rightarrow C=16x^2-8x+1+2023\)
\(\Rightarrow C=\left(4x-1\right)^2+2023\ge2023\left(\left(4x-1\right)^2\ge0\right)\)
\(\Rightarrow Min\left(C\right)=2023\)
\(D=-25x^2+50x-2023\)
\(\Rightarrow D=-\left(25x^2-50x+25\right)-1998\)
\(\Rightarrow D=-\left(5x-5\right)^2-1998\le1998\left(-\left(5x-5\right)^2\le0\right)\)
\(\Rightarrow Max\left(D\right)=1998\)
\(B=-x^2+20x+100=-\left(x^2-20x+100\right)+200=-\left(x-10\right)^2+200\le200\left(-\left(x-10\right)^2\le0\right)\)
\(\Rightarrow Max\left(B\right)=200\)
\(E=\left(2x-1\right)^2-\left(3x+2\right)\left(x-5\right)\)
\(\Rightarrow E=4x^2-4x+1-\left(3x^2-13x-10\right)\)
\(\Rightarrow E=4x^2-4x+1-3x^2+13x+10\)
\(\Rightarrow E=x^2+9x+11=x^2+9x+\dfrac{81}{4}-\dfrac{81}{4}+11\)
\(\Rightarrow E=\left(x+\dfrac{9}{2}\right)^2-\dfrac{37}{4}\ge-\dfrac{37}{4}\left(\left(x+\dfrac{9}{2}\right)^2\ge0\right)\)
\(\Rightarrow Min\left(E\right)=-\dfrac{37}{4}\)
\(F=\left(3x-5\right)^2-\left(3x+2\right)\left(4x-1\right)\)
\(\Rightarrow F=9x^2-30x+25-\left(12x^2+3x-2\right)\)
\(\Rightarrow F=-3x^2-33x+27=-3\left(x^2-10x+9\right)\)
\(\Rightarrow F=-3\left(x^2-10x+25\right)+48=-3\left(x-5\right)^2+48\le48\left(-3\left(x-5\right)^2\le0\right)\)
\(\Rightarrow Max\left(F\right)=48\)
12) Chọn đáp án để được khẳng định đúng (x + 10)2 = ?
A.x2 –10x+100 B.x2 +10x+100 C. x2 + 100 D. x2 + 20x + 100
Nếu đề là (x+10)2 thì đáp án là D nhé
nhớ like nha
a) 770:[(20x+100);x]=35
b) [(8x-14):2-2].31=341
c)25.[260-(x+110)-125=0
khó quá bạn nào giúp mình với
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tìm x biết:
a,(3x+ 1) x ( x-2000)x (3x +6000) = 0
b, (3x-1)^2 = 169
c,|2x+5|=1/2
d, x-6 / x-7= x+1 / x-3
e, (x-5)x30 / 100=20x/100 + 5
a) pt a <=> 3x+1=0 hoặc x-2000=0 hoặc 3x+6000=0
<=> x=-1/3 hoặc x=2000 hoặc x=-2000