Bài 2: Tìm x biết
1. (𝑥 − 3)² = 4𝑥 ²+ 20x + 25
2. 2x(x – 4) + 𝑥² – 16 = 0
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a) x³ - 64x = 0
x(x² - 64) = 0
x(x - 8)(x + 8) = 0
x = 0 hoặc x - 8 = 0 hoặc x + 8 = 0
*) x - 8 = 0
x = 8
*) x + 8 = 0
x = -8
Vậy x = -8; x = 0; x = 8
b) x³ - 4x² = -4x
x³ - 4x² + 4x = 0
x(x² - 4x + 4) = 0
x(x - 2)² = 0
x = 0 hoặc (x - 2)² = 0
*) (x - 2)² = 0
x - 2 = 0
x = 2
Vậy x = 0; x = 2
c) x² - 16 - (x - 4) = 0
(x - 4)(x + 4) - (x - 4) = 0
(x - 4)(x + 4 - 1) = 0
(x - 4)(x + 3) = 0
x - 4 = 0 hoặc x + 3 = 0
*) x - 4 = 0
x = 4
*) x + 3 = 0
x = -3
Vậy x = -3; x = 4
d) (2x + 1)² = (3 + x)²
(2x + 1)² - (3 + x)² = 0
(2x + 1 - 3 - x)(2x + 1 + 3 + x) = 0
(x - 2)(3x + 4) = 0
x - 2 = 0 hoặc 3x + 4 = 0
*) x - 2 = 0
x = 2
*) 3x + 4 = 0
3x = -4
x = -4/3
Vậy x = -4/3; x = 2
e) x³ - 6x² + 12x - 8 = 0
(x - 2)³ = 0
x - 2 = 0
x = 2
f) x³ - 7x - 6 = 0
x³ + 2x² - 2x² - 4x - 3x - 6 = 0
(x³ + 2x²) - (2x² + 4x) - (3x + 6) = 0
x²(x + 2) - 2x(x + 2) - 3(x + 2) = 0
(x + 2)(x² - 2x - 3) = 0
(x + 2)(x² + x - 3x - 3) = 0
(x + 2)[(x² + x) - (3x + 3)] = 0
(x + 2)[x(x + 1) - 3(x + 1)] = 0
(x + 2)(x + 1)(x - 3) = 0
x + 2 = 0 hoặc x + 1 = 0 hoặc x - 3 = 0
*) x + 2 = 0
x = -2
*) x + 1 = 0
x = -1
*) x - 3 = 0
x = 3
Vậy x = -1; x = -1; x = 3
Dòng cuối kết luận phải là \(\text{x }\in\text{ }\left\{-2;-1;3\right\}\) chứ ạ?
1,\(=4x\left(x-\dfrac{3}{2}\right)\)
2,\(=-7y^3\left[2x^2y\left(2y+x\right)+3\right]\)
3, = 4x(a-b)-6xy(a-b)
=2x(a-b)(2-3y)
4,
=3(2x+1)-(2x-5)(2x+1)
=(3-2x+5)(2x+1)
=(8-2x)(2x+1)
=2(4-x)(2x+1)
a) \(\sqrt{x}=3\left(x\ge0\right)\Leftrightarrow x=9\)
b) \(\sqrt{x}=\sqrt{5}\left(x\ge0\right)\Leftrightarrow x=5\)
c) \(\sqrt{x}=0\left(x\ge0\right)\Leftrightarrow x=0\)
d) \(\sqrt{x}=-2\left(x\ge0\right)\Leftrightarrow x=\varnothing\)
e) \(\sqrt{x-2}=3\left(x\ge0\right)\Leftrightarrow x-2=9\Leftrightarrow x=11\)
g) \(\sqrt{2x-1}=5\left(x\ge0\right)\Leftrightarrow2x-1=25\Leftrightarrow2x=26\Leftrightarrow x=13\)
h) \(\sqrt{x-3}=0\left(x\ge0\right)\Leftrightarrow x-3=0\Leftrightarrow x=3\)
a: \(\sqrt{x}=3\)
nên x=9
b: \(\sqrt{x}=\sqrt{5}\)
nên x=5
c: \(\sqrt{x}=0\)
nên x=0
d: \(\sqrt{x}=-2\)
nên \(x\in\varnothing\)
e: \(\sqrt{x}-2=3\)
\(\Leftrightarrow\sqrt{x}=5\)
hay x=25
g: \(\sqrt{2x}-1=5\)
\(\Leftrightarrow2x=36\)
hay x=18
h: Ta có: \(\sqrt{x}-3=0\)
nên x=9
a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)
mà -3<x<30
nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)
b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)
mà -16<=x<20
nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)
c: \(\Leftrightarrow x-1+4⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)
d: \(\Leftrightarrow2x+4-5⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-1;-3;3;-7\right\}\)
\(A=\sqrt{x^2-4x+25}=\sqrt{\left(x-2\right)^2+21}\)
Ta có : \(\left(x-2\right)^2\ge0\) => \(\left(x-2\right)^2+21\ge21\left(\forall x\right)\) => \(\sqrt{\left(x-2\right)^2+21}\ge\sqrt{21}\left(\forall x\right)\)
Dấu " = " xảy ra \(\Leftrightarrow\) \(\sqrt{\left(x-2\right)^2}=0\)
\(\Leftrightarrow\) \(x-2=0\)
\(\Leftrightarrow\) x = 2
Vậy giá trị nhỏ nhất của A là : \(\sqrt{21}\) khi x = 2
\(B=\sqrt{x^2-6x+30}=\sqrt{\left(x-3\right)^2+21}\)
Vì \(\sqrt{\left(x-3\right)^2}\ge0\left(\forall x\right)\)=> \(\sqrt{\left(x-3\right)^2+21}\ge\sqrt{21}\left(\forall x\right)\)
Dấu " = " xảy ra \(\Leftrightarrow\) \(\sqrt{\left(x-3\right)^2}=0\)
\(\Leftrightarrow\) \(x-3=0\)
\(\Leftrightarrow\) \(x=3\)
Vậy giá trị nhỏ nhất của B là : \(\sqrt{21}\) khi x = 3
\(D=\sqrt{x^2-4x+7}+\sqrt{2}=\sqrt{\left(x-2\right)^2+3}+\sqrt{2}\)
Vì
\(a,\left(x+2\right)^2+\left(x+3\right)^2-2\left(x-2\right)\left(x-3\right)=19\\ \Leftrightarrow x^2+4x+4+x^2+6x+9-2x^2+10x-12=19\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\\ b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-5\right)=15\\ \Leftrightarrow x^3+8-x^3+5x=15\\ \Leftrightarrow5x=7\\ \Leftrightarrow x=\dfrac{7}{5}\\ c,\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\\ \Leftrightarrow x^3-3x^2+3x+1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=8\\ \Leftrightarrow x=\dfrac{8}{9}\)
a. (x + 2)2 + (x + 3)2 - 2(x - 2)(x - 3) = 19
<=> (x2 + 4x + 4) + (x2 + 6x + 9) - (2x + 4)(x - 3) = 19
<=> x2 + 4x + 4 + x2 + 6x + 9 - 2x2 + 6x - 4x + 12 = 19
<=> x2 + x2 - 2x2 + 4x + 6x + 6x - 4x + 9 + 4 + 12 - 19 = 0
<=> 12x + 6 = 0
<=> 6(2x + 1) = 0
<=> 2x + 1 = 0
<=> 2x = -1
<=> x = \(\dfrac{-1}{2}\)
Lời giải:
1.
$(x-3)^2=4x^2+20x+25=(2x+5)^2$
$\Leftrightarrow (x-3)^2-(2x+5)^2=0$
$\Leftrightarrow (x-3-2x-5)(x-3+2x+5)=0$
$\Leftrightarrow (-x-8)(3x+2)=0$
$\Leftrightarrow -x-8=0$ hoặc $3x+2=0$
$\Leftrightarrow x=-8$ hoặc $x=-\frac{2}{3}$
2.
$2x(x-4)+x^2-16=0$
$\Leftrightarrow 2x(x-4)+(x-4)(x+4)=0$
$\Leftrightarrow (x-4)(2x+x+4)=0$
$\Leftrightarrow (x-4)(3x+4)=0$
$\Leftrightarrow x-4=0$ hoặc $3x+4=0$
$\Leftrightarrow x=4$ hoặc $x=-\frac{4}{3}$