tìm x biết:

(3x-1) [- 1/2x+5]=0

1/4+1/3:(2x-1)=-5

[2x+3/5]² - 9/25=0

-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6

[x+1/2]x [2/3-2x]=0

17/2-|2x-3/4|=-7/4

2/3x-1/2x =5/12

(x+1/5)²+17/25=26/25

[x.44/7+3/7].11/5-3/7=-2

3[3x-1/2]+1/9=0

Toán lớp 6Tìm x

 Trả lời  Câu hỏi tương tự

Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !

3(3x-1/2)^3+1/9=0

a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)

=>\(\left(x-2\right)\left(x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)

mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)

nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)

d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)

=>\(2^x\left(1+2+2^2+2^3\right)=120\)

=>\(2^x\cdot15=120\)

=>\(2^x=8\)

=>x=3

e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)

=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)

=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)

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 \(x=2\)

#\(Toru\)

`a)(2x-1)^2+(x+3)^2-5(x-7)(x+7)`

`=4x^2-4x+1+x^2+6x+9-5(x^2-49)`

`=5x^2-5x^2-4x+6x+1+9+245`

`=2x+255`

`b)(x-2)(x^2+2x+4)-(25+x^3)`

`=x^3-8-x^3-25=-33`

Lời giải:

a. 

$(2x-1)^2+(x+3)^2-5(x-7)(x+7)$

$=4x^2-4x+1+(x^2+6x+9)-5(x^2-49)$

$=5x^2+2x+10-(5x^2-245)=2x+255$

b.

$(x-2)(x^2+2x+4)-(25+x^3)=(x^3-2^3)-(25+x^3)$

$=-8-25=-33$

a: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)\)

\(=4x^2-4x+1+x^2+6x+9-5x^2+245\)

\(=2x+255\)

b: \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x^3+25\right)\)

\(=x^3-8-x^3-25\)

=-33

mk ko bt 123

Suy ra (2x-4)-(3x-3×5)=1 Suy ra(2x-4)-3x+15=1 Suy ra 2x-4-3x+15=1 Suy ra (2x-3x)+(15-4)=1 -1x+11=1 1-11=-1x -1x=-10 X=10

x-3)^2-2(2x-7)(x-3)=0

Cái này dễ mà, nên tự làm đi 😂😂😂

a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow-\left(2x-5\right)\left(2x+7\right)=\left(2x-5\right)\left(2x+5\right)\)

\(\Leftrightarrow-2x-7=2x+5\)

\(\Leftrightarrow-4x=12\Leftrightarrow x=-3\)

\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right].\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\7x+23=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\frac{23}{7}\end{matrix}\right.\)

b/\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow4\left(4x^2+28x+49\right)-9\left(x^2+6x+9\right)=0\)

\(\Leftrightarrow16x^2+112x+196-9x^2-36x-81=0\)

\(\Leftrightarrow7x^2+76x+115=0\)

\(\Leftrightarrow x=\frac{-38\pm3\sqrt{71}}{7}\)

Vậy ..