Tìm x A)4x²-25-(x+7)(2x-5)
HAKED BY PAKISTAN 2011
Tìm x:
a) 4x^2 - 25 - ( 2x - 5) .( 2x + 7) = 0
b) x^3 + 27 + ( x+3). ( x -9) = 0
a) 4x^2 - 25 - ( 2x - 5) .( 2x + 7) = 0
<=>4x2-25-(4x2+14x-10x-35)=0
<=>4x2-25-4x2-14x+10x+35= 0
<=>-4x+10= 0
<=>x= 5/2
b) x^3 + 27 + ( x+3). ( x -9) = 0
<=>x3+33+(x+3)(x-9)=0
<=>(x+3)(x2-3x+9)+(x+3)(x-9)=0
<=>(x+3)(x2-3x+9+x-9) =0
<=>(x+3)(x2-2x)=0
<=>(x+3)(x-2)x= 0
<=>x=-3 hoặc x=2 hoặc x=2
tìm x :
(5-2x)x(2x+7)=4x^2-25
\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)=\left(2x+5\right)\left(2x-5\right)\)
\(\Leftrightarrow-\left(2x-5\right).\left(2x+7\right)-\left(2x+5\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(-7-2x-2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(-12-4x\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\-12-4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
Vậy x=5/2 hoặc x=-3
Tìm x thuộc Z
a. (x-200).(x-5)=0
b. 2x-50+3x-7=4x+120
c. -5.(x-3)-7.(5-x)=25
a;x=200 /x=5|b;x=60|c;x=18/x=-13...tick cho mik nhé
Tìm x, biết:
a) 4x2-12x=-9
b) (5-2x)(2x+7)=4x2-25
c) x3+27+(x+3)(x-9)=0
d) 4(2x+7)2-9(x+3)2=0
tìm x:
\(\sqrt{x^2+x+1}=1\)
\(\sqrt{x^2+1}=-3\)
\(\sqrt{x^2-10x+25}=7-2x\)
\(\sqrt{2x+5}=5\)
\(\sqrt{x^2-4x+4}-2x+5=0\)
√(x² + x + 1) = 1
⇔ x² + x + 1 = 1
⇔ x² + x = 0
⇔ x(x + 1) = 0
⇔ x = 0 hoặc x + 1 = 0
*) x + 1 = 0
⇔ x = -1
Vậy x = 0; x = -1
--------------------
√(x² + 1) = -3
Do x² ≥ 0 với mọi x
⇒ x² + 1 > 0 với mọi x
⇒ x² + 1 = -3 là vô lý
Vậy không tìm được x thỏa mãn yêu cầu
--------------------
√(x² - 10x + 25) = 7 - 2x
⇔ √(x - 5)² = 7 - 2x
⇔ |x - 5| = 7 - 2x (1)
*) Với x ≥ 5, ta có
(1) ⇔ x - 5 = 7 - 2x
⇔ x + 2x = 7 + 5
⇔ 3x = 12
⇔ x = 4 (loại)
*) Với x < 5, ta có:
(1) ⇔ 5 - x = 7 - 2x
⇔ -x + 2x = 7 - 5
⇔ x = 2 (nhận)
Vậy x = 2
--------------------
√(2x + 5) = 5
⇔ 2x + 5 = 25
⇔ 2x = 20
⇔ x = 20 : 2
⇔ x = 10
Vậy x = 10
-------------------
√(x² - 4x + 4) - 2x +5 = 0
⇔ √(x - 2)² - 2x + 5 = 0
⇔ |x - 2| - 2x + 5 = 0 (2)
*) Với x ≥ 2, ta có:
(2) ⇔ x - 2 - 2x + 5 = 0
⇔ -x + 3 = 0
⇔ x = 3 (nhận)
*) Với x < 2, ta có:
(2) ⇔ 2 - x - 2x + 5 = 0
⇔ -3x + 7 = 0
⇔ 3x = 7
⇔ x = 7/3 (loại)
Vậy x = 3
1)
\(\Leftrightarrow x^2+x+1=1^2=1\\ \Leftrightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
2) Do \(x^2+1>0\forall x\) nên \(x\in\varnothing\)
3)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=7-2x\\ \Leftrightarrow\left|x-5\right|=7-2x\)
Nếu \(x\ge5\) thì
\(\Leftrightarrow x-5-7+2x=0\\ \Leftrightarrow3x-12=0\\ \Leftrightarrow3x=12\\ \Rightarrow x=4\)
=> Loại trường hợp này
Nếu \(x< 5\) thì
\(\Leftrightarrow5-x-7+2x=0\\ \Leftrightarrow x-2=0\\ \Rightarrow x=2\)
=> Nhận trường hợp này
Vậy x = 2
4)
\(\Leftrightarrow2x+5=5^2=25\\ \Leftrightarrow2x=25-5=20\\ \Rightarrow x=\dfrac{20}{2}=10\)
5)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}-2x+5=0\\ \Leftrightarrow\left|x-2\right|-2x+5=0\)
Nếu \(x\ge2\) thì
\(\Leftrightarrow x-2-2x+5=0\\ \Leftrightarrow3-x=0\\ \Rightarrow x=3\)
=> Nhận trường hợp này
Nếu \(x< 2\) thì
\(\Leftrightarrow2-x-2x+5=0\\ \Leftrightarrow7-3x=0\\ \Leftrightarrow3x=7\\ \Rightarrow x=\dfrac{7}{3}\)
=> Loại trường hợp này
Vậy x = 3
Tìm x biết : \(\left(5-2x\right)\left(2x+7\right)-4x^2-25=0\)
\(\left(5x-2\right)\left(2x+7\right)-4x^2-25=0\)
\(10x+35-4x^2-14x-4x^2+25=0\)
\(-4x+60-8x^2=0\)
\(-4\left(2x^2+x-15\right)=0\)
\(-4\left(2x^2+6x-5x-15\right)=0\)
\(-4\left(2x-5\right)\left(x+3\right)=0\)
=> \(x\) ∈ \(\left\{\dfrac{5}{2};-3\right\}\)
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
Bài 1.khai triển HĐT
a,(3x-4)^2 b,(1+4x)^2 c,(2x+3)^3
d,(5-2x)^3 e,49x^2-25 f,1/25-81y^2
Bài 2.Tìm x biết:Viết đầy đủ
a,(x-5)^2-(x+7)(x-7)=8 b,(2x+5)^2-4(x+1)(x-1)=10
Bài 3.Tìm GTLN,GTNN của các biểu thức sau
a,A=x^2-6x+19 b,B=-x^2+8x-20
c,C=4x^2+12x+100 d,D=25+4x-x^2
Bài 1.
\(a, (3x-4)^2\)
\(=\left(3x\right)^2-2\cdot3x\cdot4+4^2\)
\(=9x^2-24x+16\)
\(b,\left(1+4x\right)^2\)
\(=1^2+2\cdot1\cdot4x+\left(4x\right)^2\)
\(=16x^2+8x+1\)
\(c,\left(2x+3\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=8x^3+36x^2+54x+27\)
\(d,\left(5-2x\right)^3\)
\(=5^3-3\cdot5^2\cdot2x+3\cdot5\cdot\left(2x\right)^2-\left(2x\right)^3\)
\(=125-150x+60x^2-8x^3\)
\(e,49x^2-25\)
\(=\left(7x\right)^2-5^2\)
\(=\left(7x-5\right)\left(7x+5\right)\)
\(f,\dfrac{1}{25}-81y^2\)
\(=\left(\dfrac{1}{5}\right)^2-\left(9y\right)^2\)
\(=\left(\dfrac{1}{5}-9y\right)\left(\dfrac{1}{5}+9y\right)\)
Bài 2.
\(a,\left(x-5\right)^2-\left(x+7\right)\left(x-7\right)=8\)
\(\Rightarrow x^2-2\cdot x\cdot5+5^2-\left(x^2-7^2\right)=8\)
\(\Rightarrow x^2-10x+25-\left(x^2-49\right)=8\)
\(\Rightarrow x^2-10x+25-x^2+49=8\)
\(\Rightarrow\left(x^2-x^2\right)-10x=8-25-49\)
\(\Rightarrow-10x=-66\)
\(\Rightarrow x=\dfrac{33}{5}\)
\(b,\left(2x+5\right)^2-4\left(x+1\right)\left(x-1\right)=10\)
\(\Rightarrow\left(2x\right)^2+2\cdot2x\cdot5+5^2-4\left(x^2-1^2\right)=10\)
\(\Rightarrow4x^2+20x+25-4x^2+4=10\)
\(\Rightarrow\left(4x^2-4x^2\right)+20x=10-25-4\)
\(\Rightarrow20x=-19\)
\(\Rightarrow x=\dfrac{-19}{20}\)
#\(Toru\)
Bài 1
a) (3x - 4)²
= (3x)² - 2.3x.4 + 4²
= 9x² - 24x + 16
b) (1 + 4x)²
= 1² + 2.1.4x + (4x)²
= 1 + 8x + 16x²
c) (2x + 3)³
= (2x)³ + 3.(2x)².3 + 3.2x.3² + 3³
= 8x³ + 36x² + 54x + 27
d) (5 - 2x)³
= 5³ - 3.5².2x + 3.5.(2x)² - (2x)³
= 125 - 150x + 60x² - 8x³
e) 49x² - 25
= (7x)² - 5²
= (7x - 5)(7x + 5)
f) 1/25 - 81y²
= (1/5)² - (9y)²
= (1/5 - 9y)(1/5 + 9y)
Bài 3.
\(a,A=x^2-6x+19\)
\(=x^2-6x+9+10\)
\(=\left(x^2-2\cdot x\cdot3+3^2\right)+10\)
\(=\left(x-3\right)^2+10\)
Ta thấy: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-3\right)^2+10\ge10\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: \(Min_A=10\) khi \(x=3\)
\(b,B=-x^2+8x-20\)
\(=-x^2+8x-16-4\)
\(=-\left(x^2-8x+16\right)-4\)
\(=-\left(x^2-2\cdot x\cdot4+4^2\right)-4\)
\(=-\left(x-4\right)^2-4\)
Ta thấy: \(\left(x-4\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-4\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-4\right)^2-4\le-4\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy \(Max_B=-4\) khi \(x=4\)
\(c,C=4x^2+12x+100\)
\(=4x^2+12x+9+91\)
\(=\left[\left(2x\right)^2+2\cdot2x\cdot3+3^2\right]+91\)
\(=\left(2x+3\right)^2+91\)
Ta thấy: \(\left(2x+3\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+3\right)^2+91\ge91\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow2x+3=0\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy \(Min_C=91\) khi \(x=\dfrac{-3}{2}\)
\(d,D=25+4x-x^2\)
\(=-x^2+4x-4+29\)
\(=-\left(x^2-2\cdot x\cdot2+2^2\right)+29\)
\(=-\left(x-2\right)^2+29\)
Ta thấy: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2+29\le29\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy \(Max_D=29\) khi \(x=2\)
#\(Toru\)
Tìm x, biết
4x2 - 25 - (2x - 5) (2x+7) = 0
4x2 - 25 - (2x - 5)(2x + 7) = 0
<=> (2x)2 - 52 - (2x - 5)(2x + 7) = 0
<=> (2x - 5)(2x + 5) - (2x - 5)(2x + 7) = 0
<=> (2x - 5)(2x + 5 - 2x - 7) = 0
<=> (2x - 5) . (-2) = 0
<=> 2x - 5 = 0
<=> 2x = 5
<=> x = 5/2