Bài 22: Tìm x, biết.
a, 25x2-9=0
b, (x+4)2-(x+1) (x-1)
c, (2x-1)2+(x+3)2-5(x+7) (x-7)=0
Mọi người trình bày đầy đủ hộ e với!
Bài 3: Giải các phương trình sau:
a, 2x3 - 50x = 0
b, 2x (3x - 5) - (5 - 3x)
c, 9(3x - 2) = x(2 - 3x)
d, (2x - 1)2 - 25 = 0
e, 25x2 - 2 = 0
f, x2 - 25 = 6x - 9
g, 5x(x - 3) - 2x + 6 = 0
h, 3x(x - 7) - 2(x - 7) = 0
i, 7x2 - 28 = 0
j, (2x + 1) + x(2x + 1) = 0
k, (x + 2)2 - (x - 2)(x + 2) = 0
l, x3 + 5x2 - 4x - 20 = 0
m, x2 - 25 + 2(x + 5) = 0
n, x3 - 3x + 2 = 0
o, x2 - 6x + 8 = 0
p, x2 - 5x - 14 = 0
q, (x - 2)2 - (x - 3)(x + 3) = 6
r, (2x - 1)2 - (2x + 5)(2x - 5) = 18
Bài 4: Tìm x, biết.
a) 4x(x - 7) - 4x2 = 56
b) 12x(3x - 2) - (4 - 6x) = 0
c) 4(x - 5) - (5 - x)2 = 0
d) x(x +1) - x(x - 3) = 0
e) - 6x + 8 = 0 f) 2 + 2x + = 0
c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
alo mọi người giúp mình nha, mk mới học nên chẳng biết bài tập tìm x biết:
a) (10x+9)*x-(5x-1)*(2x+3)=8
b)(3x-5)*(7-5x)+(5x+2)(3x-2)-2=0
mọi người ghi cụ thể hộ mk nhé
a: \(\Rightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)
\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)
=>-4x=5
hay x=-5/4
b: \(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
=>42x=41
hay x=41/42
`a)(10x+9)x-(5x-1)(2x+3)=8`
`<=>10x^2+9x-10x^2-15x+2x+3=8`
`<=>-4x=5`
`<=>x=-5/4` Vậy `S={-5/4}`
`b)(3x-5)(7-5x)+(5x+2)(3x-2)-2=0`
`<=>21x-15x^2-35+25x+15x^2-10x+6x-4-2=0`
`<=>42x=41`
`<=>x=41/42` Vậy `S={41/42}`
a: ⇒10x2+9x−(10x2+15x−2x−3)=8⇒10x2+9x−(10x2+15x−2x−3)=8
⇔10x2+9x−10x2−13x+3=8⇔10x2+9x−10x2−13x+3=8
=>-4x=5
hay x=-5/4
b: ⇔21x−15x2−35+25x+15x2−10x+6x−4−2=0⇔21x−15x2−35+25x+15x2−10x+6x−4−2=0
=>42x=41
hay x=41/42
Tìm số tự nhiên x biết:
2+(x+3)=7
5+(3+x)=10
(4+x)+1=7
(x+5)+3=9
(x-1)-4=7
4-(6-x)=1
Giúp e vs ạ(Vui lòng trình bày nguyên bài)
`@` `\text {Ans}`
`\downarrow`
`2+(x+3)=7`
`\Rightarrow x+3=7-2`
`\Rightarrow x+3=5`
`\Rightarrow x=5-3`
`\Rightarrow x=2`
`5+(3+x)=10`
`\Rightarrow 3+x=10-5`
`\Rightarrow 3+x=5`
`\Rightarrow x=5-3`
`\Rightarrow x=2`
`(4+x)+1=7`
`\Rightarrow 4+x=7-1`
`\Rightarrow 4+x=6`
`\Rightarrow x=6-4`
`\Rightarrow x=2`
`(x+5)+3=9`
`\Rightarrow x+5=9-3`
`\Rightarrow x+5=6`
`\Rightarrow x=6-5`
`\Rightarrow x=1`
`(x-1)-4=7`
`\Rightarrow x-1=7+4`
`\Rightarrow x-1=11`
`\Rightarrow x=11+1`
`\Rightarrow x=12`
`4-(6-x)=1`
`\Rightarrow 6-x=4-1`
`\Rightarrow 6-x=3`
`\Rightarrow x=6-3`
`\Rightarrow x=3`
\(2+\left(x+3\right)=7\)
\(\Rightarrow2+x+3=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(5+\left(3+x\right)=10\)
\(\Rightarrow5+3+x=10\)
\(\Rightarrow x+8=10\)
\(\Rightarrow x=2\)
\(\left(4+x\right)+1=7\)
\(\Rightarrow4+x+1=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(\left(x+5\right)+3=9\)
\(=x+5+3=9\)
\(\Rightarrow x+8=9\)
\(\Rightarrow x=1\)
\(\left(x-1\right)-4=7\)
\(\Rightarrow x-1-4=7\)
\(\Rightarrow x-5=7\)
\(\Rightarrow x=12\)
\(4-\left(6-x\right)=1\)
\(\Rightarrow4-6-x=1\)
\(\Rightarrow-2-x=1\)
\(\Rightarrow x=-3\)
Giải bất phương trình:
a, 2x ² -4 < 2( x+1) ²
b, (x-3) ² - 5 < -6x
c, x ² - 4x +3 <0
MỌI NGƯỜI TRẢ LỜI ĐẦY ĐỦ VÀ CHI TIẾT NHÉ!
a,
\(\Leftrightarrow\left(\left(2x^2-4\right)-2\left(x+1\right)^2\right)< 0\)
\(\Leftrightarrow2x^2-4-2\left(x^2+2x+1\right)< 0\)
\(\Leftrightarrow2x^2-4-2x^2-4x-2< 0\)
\(\Leftrightarrow-4x-6< 0\)
\(\Rightarrow x+\dfrac{3}{2}>0\)
\(\Rightarrow x>-\dfrac{3}{2}\)
\(x\in\left\{-\dfrac{3}{2};\infty\right\}\)
b/
\(\Leftrightarrow\left(x-3\right)^2-5+6x< 0\)
\(\Leftrightarrow x^2-6x+9-5+6x< 0\)
\(\Leftrightarrow x^2+4< 0\) ( điều này vô lý vì không có giá trị nào của x khiến x^2+4<0)
từ trên suy ra:
không có giá trị nào của x để pt này đúng .
c, đưa các hệ số vào công thức bậc 2 ( áp dụng ct bậc 2):
có: \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2.a}\)
Với : a=1 ; b=-4 ; c =3
Ta có:
\(x=\dfrac{-1.-4\pm\sqrt{-4^2-4.1.3}}{2.1}\)
\(\Leftrightarrow x=\dfrac{4\pm\sqrt{4}}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{4+2}{2}\\x_2=\dfrac{4-2}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=3\\x_2=1\end{matrix}\right.\)
tìm khoảng bđt bằng parabol :
có một dấu bất đẳng thức
\(\Rightarrow1< x< 3\)
9. Tìm x, y biết.
a) |x-2|+|2x+y|≤0
b) |x-2|+|x-5|≤3
a) |x-2|+|2x+y|≤0
|x-2| ≥ 0
|2x+y|≥ 0
mà |x-2|+|2x+y|≤0⇒|x-2|+|2x+y|=0
⇒|x-2|=0⇒x=2
|2x+y|=0⇒|2.2+y|=0⇒|4+y|=0⇒y=-4
vậy (x,y)={(2;-4)}
Bài 4: Tìm x:
1) x2 - 9x = 0 2) x(x - 4) – x2 = 7 3) 3x + 2(x – 5) = 5
4) 25x2 - 1 = 0 5) 3x(x - 2) - 5(x - 2) = 0 6) 3x(x - 7) + 4(x – 7) = 0
7) 4x2 – 9 = 0 8) 10x(x - 4) + 2x - 8 = 0 9) x(2x - 5) - 2x2 = 0
10) 2x2 – 4x = 0 11) 2x(3 - 4x) + 3(4x - 3) = 0 12) 2x (x – 5) – 2x2 = 3
mọi người giúp mình vs chiều 1g mình thi rồi! cảm ơn!
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)
2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)
3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)
4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)
5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)
6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)
7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)
8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)
9) \(x\left(2x-5\right)-2x^2=0\Rightarrow x\left(2x-5-2x=0\right)\Rightarrow x=0\)
10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)
11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)
12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)
Giải giúp mình mấy bài tìm x nhé (trình bày đầy đủ nha)
a) 6x+x=511:59+31
b)7x-x=521:519+3.22-70
c)9x-1=9
d)4x=64
e)x4=16
a , 6x + x=5¹¹ : 5⁹ + 3¹
7x= 5²+ 3¹
7x = 25+3
7x = 28
x = 4
Tìm x biết.
a)(x+2)3-x2(x+6)=0
b) (2x+3)3-8x(x-1)(x+1)=9x(4x-3)
c)(2-x)3+(2+x)3-12x(x+1)=0
a) \(\left(x+2\right)^3-x^2\left(x+6\right)=0\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=0\)
\(\Leftrightarrow12x+8=0\)
\(\Leftrightarrow12x=-8\)
\(\Leftrightarrow x=-\dfrac{8}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
b) \(\left(2x+3\right)^3-8x\left(x+1\right)\left(x-1\right)=9x\left(4x-3\right)\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x\left(x^2-1\right)=36x^2-27x\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3+8x=36x^2-27x\)
\(\Leftrightarrow8x^3-8x^3+36x^2-36x^2+54x+27x+8x+27=0\)
\(\Leftrightarrow89x+27=0\)
\(\Leftrightarrow x=-\dfrac{27}{89}\)
c) \(\left(2-x\right)^3+\left(2+x\right)^3-12x\left(x+1\right)=0\)
\(\Leftrightarrow8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x^2-12x=0\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(6x^2+6x^2-12x^2\right)-\left(12x-12x\right)+12x+\left(8+8\right)=0\)
\(\Leftrightarrow12x+16=0\)
\(\Leftrightarrow x=-\dfrac{16}{12}\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
`#040911`
`a)`
`(x + 2)^3 - x^2(x + 6) = 0`
`<=> x^3 + 6x^2 + 12x + 8 - x^3 - 6x^2 = 0`
`<=> (x^3 - x^3) + (6x^2 - 6x^2) + 12x = 0`
`<=> 12x = 0`
`<=> x = 0`
Vậy, `x = 0.`
`b)`
`(2x + 3)^3 - 8x(x - 1)(x + 1) = 9x(4x - 3)`
`<=> 8x^3 + 36x^2 + 54x + 27 - 8x(x^2 - 1) = 36x^2 - 27x`
`<=> 8x^3 + 36x^2 + 54x + 27 - 8x^3 + 8x - 36x^2 + 27x = 0`
`<=> (8x^3 - 8x^3) + (36x^2 - 36x^2) + (54x + 8x + 27x) + 27 = 0`
`<=> 89x + 27 = 0`
`<=> 89x = -27`
`<=> x = -27/89`
Vậy, `x = -27/89`
`c)`
`(2 - x)^3 + (2 + x)^3 - 12x(x + 1) = 0`
`<=> 8 - 12x + 6x^2 - x^3 + 8 + 12x + 6x^2 + x^3 - 12x^2 - 12x = 0`
`<=> (-x^3 + x^3) + (12x - 12x - 12x) + (6x^2 + 6x^2 - 12x^2) + (8 + 8)=0`
`<=> -12x + 16 = 0`
`<=> -12x = -16`
`<=> 12x = 16`
`<=> x=4/3`
Vậy, `x = 4/3.`