A) 18 x ( 16 - 16 ) = 0
B) ( x -15) x 72 =0
C) 2 x -11 =23
D) 125 + 3x = 251
E) 315 + ( 146 - x ) = 401
G) 2 x ( x + 4) + 5 = 65
CÓ TRÌNH BÀY NHÉ !
bài tìm x này tới 20 câu lận nhé và mik cần cách trình bày lun nhé chứ k pải đáp số không. Mong các bn giúp mik
1) 21+(x+23)=50
2)125+(x-100)=150
3)24+(10-x)=26
4)148+(78+x)=244
5)(x+97)-140=12
6)(x+97)-100=12
7)56-(x+14)=22
8)491-(x-83)=72
9)(x-17)-143=0
10)241+(107-x)
11)172-(x+18)=93
12)(x+72)-184=56
13)(2x+1)-7=14
14)15-(3x-1)+11=59
15)70-5(x-3)=45
16)412-15.(nhân)(27-x)=97
17)2008(2x+7)=112
18)126-17(3x-5)=7
19)(2600+6400)-3x=1200
20)123-5(x+4)=38
Cau 1)
21+(x+23)=50
X+23=50-21
X=29-23
X=6
Cau 2)
125+(x-100)=150
X-100=150-125
X=25+100
X=125
Cau 3)
24+(10-x)=26
10-x=26-24
X=10-2
X=8
a) 2x2 + 2x(5 - x)=12 d) 2(x + 5) - x2 - 5x = 0 g) (3x + 1)2 - (x+1) = 0
b) (5 - 2x)2 - 16 = 0 e) (2x - 1)2 - 4(x + 7)(x - 7) = 0 h) x2 + 7x - 8 = 0
c) 3x2 - 3x(x-2) = 36 f) (x + 4)2 - (x + 1)(x - 1) = 16 i) -2x2 +13x -15 = 0
mik cần gấp, cảm ơn mọi người.
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
Bài4:Tìm x:
a. 124 + (118 – x) = 217 b. 7x – 8 = 713 c. x- 36:18 = 12 d. (x- 36):18 = 12 e. (x-47) -115 = 0 f. 315 + (146 – x) = 401 | g. (6x – 39 ) : 3 = 201 h. 23 + 3x = 56 : 53 i. 5x + x = 39 – 311:39 j. 7x – x = 521 : 519 + 3.22 - 70 k. 7x – 2x = 617: 615 + 44 : 11 l. 0 : x = 0 | m. 3x = 9 n. 4x = 64 o. 2x = 16 p. 9x- 1 = 9 q. x4 = 16 r. 2x : 25 = 1 |
Bài4:Tínhnhanh
1. 58.75 + 58.50 – 58.25 2. 27.39 + 27.63 – 2.27 3. 128.46 + 128.32 + 128.22 4. 87.23 + 13.93 + 70.87
| 5. 12.35 + 35.182 – 35.94 6. 48.19 + 48.115 + 134.52 7. 27.121 – 87.27 + 73.34 8. 136.23 + 136.17 – 40.36 9. 125.98 – 125.46 – 52.25 | 10. 17.93 + 116.83 + 17.233 11. 5.23 + 35.41 + 64.65 12. 29.87 – 29.23 + 64.71 13. 19.27 + 47.81 + 19.20
|
b: Ta có: 7x-8=713
nên 7x=721
hay x=103
c: Ta có: x-36:18=12
nên x-2=12
hay x=14
d: Ta có: (x-36):18=12
nên x-36=216
hay x=252
a. 124 + (118 – x) = 217
118 - x = 93
- x = - 25
x = 25
b. 7x – 8 = 713
7x =721
x =103
c. x- 36:18 = 12
x- 2 =12
x =14
d. (x- 36):18 = 12
x-36 = 216
x = 252
e. (x-47) -115 = 0
x-47 = 115
x = 162
f. 315 + (146 – x) = 401
146-x =86
-x = -60
x = 60
g. (6x – 39 ) : 3 = 201
6x - 39 = 603
6x = 642
x = 107
h. 23 + 3x = 56 : 53
23 +3x = 125
3x = 102
x=34
Tìm x biết:
1,
a,3x(x+1) - 2x(x+2) = -x-1
b,2x(x-2020) - x+2020 = 0
c,(x-4)2 - 36 = 0
d,x2 + 8x - 16 = 0
e,x(x+6) - 7x - 42 = 0
f,25x2 - 16 = 0
2,
a,3x3 - 12x = 0
b,x2 + 3x - 10 = 0
Bài 1:
a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)
\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)
b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)
e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Giải phương trình dạng tích:
a.16-25x^2=0
b.(x+1)^2-4=0
c.(3x+1)^2-4x^2=0
d.(4x+1)-(x-2)^2=0
e.(2x+1)^2-(x+3)^2=0
\(a,\Leftrightarrow\left(4-5x\right)\left(4+5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1-2\right)\left(x+1+2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(3x+1-2x\right)\left(3x+1+2x\right)=0\\ \Leftrightarrow\left(x+1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{5}\end{matrix}\right.\\ d,Sửa:\left(4x+1\right)^2-\left(x-2\right)^2=0\\ \Leftrightarrow\left(4x+1-x+2\right)\left(4x+1+x-2\right)=0\\ \Leftrightarrow\left(3x+3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{5}\end{matrix}\right.\\ e,\Leftrightarrow\left(2x+1-x-3\right)\left(2x+1+x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Giải các phương trình sau:
a) x^2 – 4x + 4 = 25
b) (5 – 2x)^2 – 16 = 0
c) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15
a) \(x^2-4x+4=25\\ \Rightarrow\left(x-2\right)^2=25\\ \Rightarrow\left[{}\begin{matrix}x-2=-5\\x-2=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
b) \(\left(5-2x\right)^2-16=0\\ \Rightarrow\left(5-2x\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}5-2x=-4\\5-2x=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4,5\\0,5\end{matrix}\right.\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\\ \Rightarrow\left(x-3\right)^3-\left(x-3\right)^3+9\left(x+1\right)^2=15\\ \Rightarrow9\left(x+1\right)^2=15\\ \Rightarrow\left(x+1\right)^2=\dfrac{5}{3}\\ \Rightarrow\left[{}\begin{matrix}x+1=-\sqrt{\dfrac{5}{3}}\\x+1=\sqrt{\dfrac{5}{3}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3+\sqrt{15}}{3}\\x=\dfrac{-3+\sqrt{15}}{3}\end{matrix}\right.\)
a)\(\Leftrightarrow\)\(x^2-4x-21=0\)
\(\Leftrightarrow\)\(x^2-7x+3x-21=0\)
\(\Leftrightarrow\)\(x(x-7)+3(x-7)=0\)
\(\Leftrightarrow\)\((x-7)(x+3)=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x=7\\ x=-3 \end{array} \right.\)
b)\(\Leftrightarrow\)\((5-2x)^2-4^2=0\)
\(\Leftrightarrow\)\((5-2x-4)(5-2x+4)=0\)
\(\Leftrightarrow\)\((-2x+1)(-2x+9)=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x=\dfrac{1}{2}\\ x=\dfrac{9}{2} \end{array} \right.\)
c)\((x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15\)
\(\Leftrightarrow\)\(x^3-9x^2+27x-27-x^3+27+9x^2+18x+9-15=0\)
\(\Leftrightarrow\)\(45x-6=0\)
\(\Leftrightarrow\)\(x=\dfrac{2}{15}\)
Bài 2: Tìm x, biết :
a) 3x^2-15x=0
b) 2x(x-3)+3-x=0
c) x^3-3x^2-6x-6+8=0
d) (x-2)^2-16(5-2x)^2
cảm ơn trước ạ
\(a,\Rightarrow3x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ c,Đề.sai\\ d,Sửa:\left(x-2\right)^2-16\left(5-2x\right)^2=0\\ \Rightarrow\left[x-2-4\left(5-2x\right)\right]\left[x-2+4\left(5-2x\right)\right]=0\\ \Rightarrow\left(x-2-20+8x\right)\left(x-2+20-8x\right)=0\\ \Rightarrow\left(9x-22\right)\left(18-7x\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{9}\\x=\dfrac{18}{7}\end{matrix}\right.\)
Gi ải các phương trình sau (Đặt ẩn phụ)
a)( x2+x)2+4(x2+x)-12=0
b) (x2+2x+3)-9(x2+2x+3)+18=0
c) (x-2)(x+2)(x2-10)=72
a: Đặt \(a=x^2+x\)
Phương trình ban đầu sẽ trở thành \(a^2+4a-12=0\)
=>\(a^2+6a-2a-12=0\)
=>a(a+6)-2(a+6)=0
=>(a+6)(a-2)=0
=>\(\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
=>\(x^2+x-2=0\)(Vì \(x^2+x+6=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\forall x\))
=>\(\left(x+2\right)\left(x-1\right)=0\)
=>\(\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
b:
Sửa đề: \(\left(x^2+2x+3\right)^2-9\left(x^2+2x+3\right)+18=0\)
Đặt \(b=x^2+2x+3\)
Phương trình ban đầu sẽ trở thành \(b^2-9b+18=0\)
=>\(b^2-3b-6b+18=0\)
=>b(b-3)-6(b-3)=0
=>(b-3)(b-6)=0
=>\(\left(x^2+2x+3-3\right)\left(x^2+2x+3-6\right)=0\)
=>\(\left(x^2+2x\right)\left(x^2+2x-3\right)=0\)
=>\(x\left(x+2\right)\left(x+3\right)\left(x-1\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x+2=0\\x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=-3\\x=1\end{matrix}\right.\)
c: \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)
=>\(\left(x^2-4\right)\left(x^2-10\right)=72\)
=>\(x^4-14x^2+40-72=0\)
=>\(x^4-14x^2-32=0\)
=>\(\left(x^2-16\right)\left(x^2+2\right)=0\)
=>\(x^2-16=0\)(do x2+2>=2>0 với mọi x)
=>x2=16
=>x=4 hoặc x=-4
Bài 2: giải phương trình sau
a) \(X^4\)-\(x^2\)-2=0
b) (x+1)\(^4\)-x\(^2\)+2)\(^2\)=0
c)3x\(^2\)-2x-8=0
Bài 3: giải phương trình sau
a) x\(^3\)-0,25=0
b) x\(^4\)+2x\(^3\)+x\(^2\)=0
c) x\(^3\)-1=0
d) 6x\(^2\)-7x+2=0
Mong có người giải giùm xin kẻm ơn :>
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)