1. Tìm x
a). I x I + I x + 1 I + I x + 2 I + I x + 3 I + I x + 4 I = 5x
b). ( 2x - 5 ) - ( 3x - 7 ) = x+ 3
Bài 1 : Tìm x , biết
a ) (x+1)*(x+3)-x*(x+2)=7
b ) 2x*(3x-5)-x(6x-1)=33
Bài 2 : Rút gọn rồi tính giá trị của biểu thức
a ) A = 5x*(4x^2-2x+1)-2x*(10x^2-5x-2) với x=15
b ) B=5x*(x-4y)-4y*(y-5x) với x=-1/5 ; y=-1/2
Bài 3 : Chứng minh rằng các biểu thức sau có giá trị không phụ thuộc vào giá trị của biến số
a ) 3x-5)*(2x+11)-(2x+3)*(3x+7)
b ) ( x-5)*(2x+3)-2x*(x-3)+x+7
bài 1:
a. \((x+1)(x+3) - x(x+2)=7 \)
\(x^2+ 3x +x +3 - x^2 -2x =7\)
\(x^2+4x+3-x^2-2x=7\)
\(=> 2x+3=7\)
\(2x=4\)
\(x = 2\)
Bài 2:
a)
\((3x-5)(2x+11) -(2x+3)(3x+7) \)
\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)
\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)
\(=-65\)
\(\)
Tìm X
a) (\(\dfrac{1}{4}\) - X) ( X + \(\dfrac{2}{5}\) ) = 0
b) I 2x + 1 I +\(\dfrac{2}{3}\) = 2
c) (2x - 3 )\(^2\) = 36
d) 7\(^x\) + 2 +2 x 7\(^x\) = 357
a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)
=>\(\left|2x+1\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
c: (2x-3)2=36
=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
d: \(7^{x+2}+2\cdot7^x=357\)
=>\(7^x\cdot49+7^x\cdot2=357\)
=>\(7^x=7\)
=>x=1
a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
\(---\)
b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)
\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)
\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
\(---\)
c) \(\left(2x-3\right)^2=36\)
\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(---\)
d) \(7^{x+2}+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)
\(\Rightarrow7^x\cdot\left(49+2\right)=357\)
\(\Rightarrow7^x\cdot51=357\)
\(\Rightarrow7^x=357:51\)
\(\Rightarrow7^x=7\)
\(\Rightarrow x=1\)
I 7+5x I = 1-4x
I 4x^2 - 2x I + 1 = 2x
I x^2 - 5x + 4 I = x+4
I 4 - 3x I = 3x -4
I 1+5x I = 1 + 5x
I x^2 - 3x + 1 I = 2x-3
I x-1 I = x^2 -x
|7 + 5x| = 1 - 4x
=> \(\orbr{\begin{cases}7+5x=1-4x\left(đk:x\le\frac{1}{4}\right)\\7+5x=4x-1\left(đk:x\ge\frac{1}{4}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}7-1=-4x-5x\\7+1=4x-5x\end{cases}}\)
=> \(\orbr{\begin{cases}6=-9x\\8=-x\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{2}{3}\left(tm\right)\\x=-8\left(ktm\right)\end{cases}}\)
|4x2 - 2x| + 1 = 2x
=> |4x2 - 2x| = 2x - 1
=> \(\orbr{\begin{cases}4x^2-2x=2x-1\left(đk:x\ge\frac{1}{2}\right)\\4x^2-2x=1-2x\left(đk:x\le\frac{1}{2}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}4x^2-2x-2x+1=0\\4x^2-2x-1+2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\4x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\x^2=\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\pm\frac{1}{2}\end{cases}}\)(tm)
Vậy ...
Tìm x \(\in\)Z , biết:
a)(2x-5)+17=6
b)10-2(4-3x)=-4
c)-12+3(-x+7)=-18
d)24(3x-2)=-3
e)-45:5(3x-2x)=3
g)x(x+7)=0
h)(x+12)(x-3)=0
i)(-x+5)(3-x)=0
k)x(2+x)(7-x)=0
l)(x-1)(x+2)(-x-3)=0
a) (2x-5) + 17 = 6
2x - 5 = 6 - 17
2x - 5 = -11
2x = -11 + 5
2x = -6
x = -6 : 2
x = -3
* Các câu b→e bạn cũng làm tương tự theo trật tự như vậy là được
* Các câu từ g → l thì bạn áp dụng lí thuyết sau:
Tích của hai số bằng 0 khi một trong hai số đó bằng 0
VD : g) x(x+7)=0
⇒ hoặc là x = 0 hoặc là x+7 = 0
( Bạn làm phép tính nhớ bỏ dấu ngoặc vuông trước nhé )
b: \(\Leftrightarrow2\left(4-3x\right)=14\)
=>4-3x=7
=>3x=-3
=>x=-1
c: \(\Leftrightarrow3\left(7-x\right)=-18+12=-6\)
=>7-x=-2
=>x=9
d: \(\Leftrightarrow3x-2=-\dfrac{1}{8}\)
=>3x=15/8
=>x=5/8
e: \(\Leftrightarrow5\left(3x-2x\right)=-15\)
=>x=-3
g: =>x=0 hoặc x+7=0
=>x=0 hoặc x=-7
h: =>x+12=0 hoặc x-3=0
=>x=3 hoặc x=-12
k: =>x=0 hoặc x+2=0 hoặc 7-x=0
=>\(x\in\left\{0;-2;7\right\}\)
l: =>x-1=0 hoặc x+2=0 hoặc x+3=0
=>\(x\in\left\{1;-2;-3\right\}\)
Giải phương trình:
1, \(3x^2+6x-3=\sqrt{\dfrac{x+7}{3}}\) (2 cách khác nhau )
2, \(\left(\sqrt{3x+1}-\sqrt{x-2}\right)\left(\sqrt{3x^2+7x+2}+4\right)=4x-2\)
3, \(\sqrt{-3x-1}+\sqrt{9x^2+9x+3}=-9x^2-6x\)
4, \(\sqrt{x^2+x-6}+3\sqrt{x-1}=\sqrt{5x^2-1}\)
5, \(\left(\sqrt{x+4}+2\right)\left(x+2\sqrt{x-5}+1\right)=6x\)
6, \(\sqrt{5-x^4}-\sqrt[3]{3x^2-2}=1\)
7, \(3x^2+11+\sqrt{x-2}+\sqrt{2x+3}=14x\)
8, \(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-7}}}}=7\)
9, \(\sqrt{2x^2-1}+3x\sqrt{x^2-1}=3x^3+2x^2-9x-7\) ( với \(x>0\) )
A=3 I x-1 I -2 I 5-3x I
B=4 I x-3 I +2 I 2x -1 I +I 4-3x I
A=(3x-3)-(10-6x)
=3x-3-10+6x
=6x+3x-3-10
=9x-13
B=(4x-12)+(4x-2)+(4-3x)
=4x-12+4x-2+3-3x
=5x-11
Giải phương trình:
1. (x - 4)2 - 25 = 0
2. (x - 3)2 - (x - 1)2 = 0
3. (x2 - 4)(2x +3) = (x2 - 4)(x - 1)
4. (x2 - 1) - (x + 1)(2 - 3x) = 0
5. x3 + x2 + x + 1 = 0
6. x3 + x2 - x - 1 = 0
7. 2x3 + 3x2 + 6x + 5 = 0
8. x4 - 4x3 - 19x2 + 106x - 120 = 0
9. (x2 - 3x + 2)(x2 + 15x + 56) + 8 = 0
1 ) \(\left(x-4\right)^2-25=0\)
\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
2 ) \(\left(x-3\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-3+x-1\right)\left(x-3-x+1\right)=0\)
\(\Leftrightarrow-2\left(2x-4\right)=0\)
\(\Leftrightarrow x=2.\)
3 ) \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x+3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.\)
4 ) \(\left(x^2-1\right)-\left(x+1\right)\left(2-3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)
5 ) \(x^3+x^2+x+1=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-1.\end{matrix}\right.\)
6 ) \(x^3+x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
7 ) \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)
\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=-1.\)
8 ) \(x^4-4x^3-19x^2+106x-120=0\)
\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)
\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)
\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)
\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)
Đặt \(x^2+6x-7=t\)
\(\Leftrightarrow t\left(t-9\right)+8=0\)
\(\Leftrightarrow t^2-9t+8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)
Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)
Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)
Vậy ........
các bạn làm mấy bài tìm x này theo bảng xét dấu hộ mk
a. I 5+2x I +2x=x-3
b.I x+1 I + I x+2 I=0
c.I 3x-1 I - I 2x+3 I = 0
d. I x-1 I -I 2-x I +x=0
b ) [( 6x - 39) : 7] .4 = 12 c) 4( 3x - 4 ) - 2 = 18 d ) ( 3x - 10 ) :10 = 50
e) x- 7 = - 57 f ) x - [ 42 + (-25) = - 8 g) ( 3x - 24 ) . 73 = 2 . 74
h) x + 5 = 20 - ( 12 -7) k) I x - 5 I = 7 - ( -3) i) I x - 5 I = I 7 I
2x+1 . 22009 = 220010 10 - 2x = 25 - 3x
nik lộn
2x+1. 22009 = 22010
10 - 2x = 25 - 3x