Tìm x, y biết:
a,4x=2x+1
b,16=(x-1)4
c,x2+2y=25
\(Tìm Min : B=2x²-4x-8 C=x²-2xy+2y²+2x-10y+17 D=x²-xy+y²-2x-2y E=(x²+x-6)(x²+x+2) F=(x+1)(x+2)(x+3)(x+4) Tìm Max G= 4x-x2 H=25-x-5x2 \)
B = 2\(x^2\) - 4\(x\) - 8
B = 2(\(x^2\) - 2\(x\) + 4) - 16
B = 2(\(x-2\))2 - 16
Vì (\(x-2\))2 ≥ 0 ∀ \(x\) ⇒ 2(\(x-2\))2 ≥ 0 ∀ \(x\)
⇒ 2(\(x-2\))2 - 16 ≥ -16 ∀ \(x\)
Dấu bằng xảy ra khi (\(x-2\))2 = 0 ⇒ \(x-2=0\) ⇒ \(x=2\)
Vậy Bmin = -16 khi \(x=2\)
Tìm min của C biết:
C = \(x^2\) - 2\(xy\) + 2y2 + 2\(x\) - 10y + 17
C = (\(x^2\) - 2\(xy\) + y2) + 2(\(x\) - y) + y2 - 8y + 16 + 1
C = (\(x\) - y)2 + 2(\(x\) - y) + 1 + (y2 - 8y + 16)
C = (\(x-y+1\))2 + (y - 4)2
Vì (\(x\) - y + 1)2 ≥ 0 ∀ \(x;y\); (y - 4)2 ≥ 0 ∀ y
Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x-y+1=0\\y=4\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x-4+1=0\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-1+4\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
Vậy Cmin = 0 khi (\(x;y\)) = (3; 4)
D = \(x^2\) - \(xy\) + y2 - 2\(x\) - 2y
D=[\(x^2\)-2\(x\)\(\dfrac{y}{2}\)+(\(\dfrac{y}{2}\))2]-(2\(x\)-2\(\dfrac{y}{2}\)) +1 +(\(\dfrac{3}{4}\)y2-2.\(\dfrac{\sqrt{3}}{2}\)y .\(\sqrt{3}\) +3) - 4
D = (\(x-\dfrac{y}{2}\))2 - 2(\(x-\dfrac{y}{2}\))+ 1 + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))2 - 4
D = (\(x-\dfrac{y}{2}\) - 1)2 + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))2 - 4
Vì (\(x-\dfrac{y}{2}\) - 1)2 ≥ 0 ∀ \(x\);y và (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))2 ≥ 0 ∀ y
Vậy (\(x-\dfrac{y}{2}\) - 1)2 + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))2 - 4 ≥ - 4 ∀ \(x;y\)
Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-1=0\\\dfrac{\sqrt{3}}{2}y-\sqrt{3}=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-1=0\\\sqrt{3}.\left(\dfrac{1}{2}y-1\right)=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=1+\dfrac{1}{2}y\\\dfrac{1}{2}y=1\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=1+1\\y=1:\dfrac{1}{2}\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy Dmin = - 4 khi (\(x;y\)) =(2; 2)
Bài 5: Tìm nghiệm của các đa thức sau: Dạng 1: a) 4x + 9 b) -5x + 6 c) 7 – 2x d) 2x + 5 Dạng 2: a) ( x+ 5 ) ( x – 3) b) ( 2x – 6) ( x – 3) c) ( x – 2) ( 4x + 10 ) Dạng 3: a) x2 -2x b) x2 – 3x c) 3x2 – 4x d) ( 2x- 1)2 Dạng 4: a) x2 – 1 b) x2 – 9 c)– x 2 + 25 d) x2 - 2 e) 4x2 + 5 f) –x 2 – 16 g) - 4x4 – 25 Dạng 5: a) 2x2 – 5x + 3 b) 4x2 + 6x – 1 c) 2x2 + x – 1 d) 3x2 + 2x – 1
a) (2x + 3y)2
b) (x + \(\dfrac{1}{4}\))2
c) (x2 + \(\dfrac{2}{5}\)y) . (x2 - \(\dfrac{2}{5}\)y)
d) (2x + y2)3
e) (3x2 - 2y)2
f) (x + 4) (x2 - 4x + 16)
g) (x2 - \(\dfrac{1}{3}\)) . (x4 + \(\dfrac{1}{3}\)x2 + \(\dfrac{1}{9}\))
a) \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)
b) \(\left(x+\dfrac{1}{4}\right)^2=x^2+2\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)
c) \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4}{25}y^2\)
d) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y^2+3\cdot2x\cdot\left(y^2\right)^2+\left(y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
e) \(\left(3x^2-2y\right)^2=\left(3x^2\right)^2-2\cdot3x^2\cdot2y+\left(2y\right)^2=9x^4-12x^2y+4y^2\)
f) \(\left(x+4\right)\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)
g) \(\left(x^2-\dfrac{1}{3}\right)\cdot\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)
Tìm Min :
B=2x²-4x-8
C=x²-2xy+2y²+2x-10y+17
D=x²-xy+y²-2x-2y
E=(x²+x-6)(x²+x+2)
F=(x+1)(x+2)(x+3)(x+4)
Tìm Max
G= 4x-x2
H=25-x-5x2
\(B=2x^2-4x-8=2\left(x^2-2x-4\right)\)
\(=2\left(x^2-2x+1-5\right)\)
\(=2\left[\left(x-1\right)^2-5\right]\)
\(=2\left(x-1\right)^2-10\ge-10\)
Vậy \(B_{min}=-10\Leftrightarrow x-1=0\Leftrightarrow x=1\)
\(F=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)
Đặt \(x^2+5x+4=t\)
\(\RightarrowĐT=t\left(t+2\right)=t^2+2t+1-1\)
\(=\left(t+1\right)^2-1\ge-1\)
hay \(\left(x^2+5x+5\right)^2-1\ge-1\)
Vậy \(F_{min}=-1\Leftrightarrow x^2+5x+5=0\)
\(\Leftrightarrow x^2+5x+\frac{25}{4}-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=\frac{5}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5}{2}=\sqrt{\frac{5}{4}}\\x+\frac{5}{2}=-\sqrt{\frac{5}{4}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{5}{4}}-\frac{5}{2}\\x=-\sqrt{\frac{5}{4}}-\frac{5}{2}\end{cases}}\)
\(G=4x-x^2=-\left(x^2-4x+4-4\right)\)
\(=-\left[\left(x-2\right)^2-4\right]=-\left(x-2\right)^2+4\le4\)
Vậy \(G_{max}=4\Leftrightarrow x-2=0\Leftrightarrow x=2\)
\(H=25-x-5x^2=-5\left(x^2+\frac{x}{5}-5\right)\)
\(=-5\left(x^2+2x.\frac{1}{10}+\frac{1}{100}-\frac{501}{100}\right)\)
\(=-5\left[\left(x+\frac{1}{10}\right)^2-\frac{501}{100}\right]\)
\(=-5\left(x+\frac{1}{10}\right)^2+\frac{101}{20}\le\frac{101}{2}\)
Vậy \(H_{max}=\frac{101}{2}\Leftrightarrow x+\frac{1}{10}=0\Leftrightarrow x=-\frac{1}{10}\)
Bài 1: Cho a/b = b/c = c/a . CMR: a = b = c
Bài 2: Cho a/b+c = b/c+a = c/a+b = K . Tìm K
Bài 3: Cho 3x-2y/4 = 2z-4x/3 = 4y-3z/2 . CMR: x/2 = y/3 = z/4
Bài 4: Cho x+16/9 = y-25/16 = z+9/25. Biết 2x^3-1 = 15 . Tìm x, y ,z
( Các dấu / là phân số)
b1 \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)
\(\frac{a}{b}=1\Rightarrow a=b;\frac{b}{c}=1\Rightarrow b=c;\frac{c}{a}=1\Rightarrow c=a\)
\(\Rightarrow a=b=c\)
b2 \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}=k\)
=> \(k=\frac{1}{2}\)
Bài 3: Phân tích các đa thức sau thành nhân tử:
a) x2 + 10x + 25. b) 8x - 16 - x2
c) x3 + 3x2 + 3x + 1 d) (x + y)2 - 9x2
e) (x + 5)2 – (2x -1)2
Bài 4: Tìm x biết
a) x2 – 9 = 0 b) (x – 4)2 – 36 = 0
c) x2 – 10x = -25 d) x2 + 5x + 6 = 0
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2
Cho 3x-2y/4 = 2z-4x/3 = 4y-3z/2. CMR: x/2 = y/3 = z/4
Cho x+16/9 = y-25/16 = z+9/25 và 2x^3-1 = 15. Tìm x, y, z
1) Cho 3x-2y/4=2z-4x/3=ay-3z/2.chứng tỏ x/2=y/3=z/4
2) tìm x,y,z biết x+16/9=y-25/16=z+9/25 và (2x^3)-1=15
3) cho a/b=c/d chứng tỏ (a-b/c-d)^2=ab/cd và (a+b/c+d)^3=a^3-b^3/c^3-d^3
4) Cmr:
10^n-18n-1 chia het cho 27
27^8-3^21 chia het cho 26
8^12-2^33-2^30 chia het cho 53
Bài 1. Phân tích đa thức 2x – 4y thành nhân tử được kết quả là:
A.2(x – 2y) B. 2( x + y) C. 4(2x – y) D. 2(x + 2y)
Bài 2. Phân tích đa thức 4x2 – 4xy thành nhân tử được kết quả là:
A.4(x2 – xy) B. x(4x – 4y) C. 4x(x – y) D. 4xy(x – y)
Bài 3. Tại x = 99 giá trị biểu thức x2 + x là:
A.990 B. 9900 C. 9100 D. 99000
Bài 4. Các giá trị của x thỏa mãn biểu thức x2 – 12x = 0 là:
A.x = 0 B. x = 12 C. x = 0 và x = 12 D. x = 11
Giúp mik với mik cảm ơn
tìm x,y biết
a) -2x (10x -3 ) + 5x (4x + 1) = 25
b)y (5 - 2y) + 2y (y - 1) = 15
c) x (x + 1) - (x+ 1) = 35
a, \(-2x\left(10x-3\right)+5x\left(4x+1\right)=25\)
\(\Rightarrow-20x^2+6x+20x^2+5x=25\)
\(\Rightarrow11x=25\Rightarrow x=\dfrac{25}{11}\)
b, \(y\left(5-2y\right)+2y\left(y-1\right)=15\)
\(\Rightarrow5y-2y^2+2y^2-2y=15\)
\(\Rightarrow3y=15\Rightarrow y=5\)
c, \(x\left(x+1\right)-\left(x+1\right)=35\)
\(\Rightarrow\left(x+1\right).\left(x-1\right)=35\)
\(\Rightarrow x^2-1=35\Rightarrow x^2=36\Rightarrow x=\pm6\)
Chúc bạn học tốt!!!