Tìm x:
a) 2^x x 4 = 128
b) ( 2x + 1)^3 = 27
c) ( 2x+1)^3 = 125
d) x^15 = x
e) ( x^5)^10 = x
f) ( x-2)^2 = (x- 2)
g) ( x-3)^2 = ( x-3)^6
Tìm x:
a) (2x - 3)(6 - 2x) = 0
b) \(5\dfrac{4}{7}:x=13\)
c) 2x - \(\dfrac{3}{7}\) = \(6\dfrac{2}{7}\)
d) \(\dfrac{x}{5}\) + \(\dfrac{1}{2}\) = \(\dfrac{6}{10}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
f) \(\dfrac{x-12}{4}=\dfrac{1}{2}\)
g) \(2\dfrac{1}{4}\).\(\left(x-7\dfrac{1}{3}\right)=1,5\)
h) \(\left(4,5-2x\right).1\dfrac{4}{7}=\dfrac{11}{14}\)
i) \(\dfrac{2}{3}\left(x-25\%\right)=\dfrac{1}{6}\)
k) \(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)
⇒\(x-12=2\)
\(x=2+12\)
x = 14
g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)
\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\)
\(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)
\(x-\dfrac{22}{3}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+\dfrac{22}{3}\)
\(x=8\)
Tìm x:
a) 165-(35x+3).19=13
b) 3.(x+7)-15=27
c) 96: (mở ngoặc vuông) 3.(x+1) (đóng ngoặc vuông)=2
d) (120-5x)+10 mủ 2.2:5=2 mủ 2.15
e) (x-1)+(x-2)+(x-3)+(x-4)=30
g) (x+1)+(2x+2)+(3x+3)+(4x+4)=70
Lưu ý: Trong câu c, do mik ko có dấu ngoặc vuông nên mik phải viết ra. Tiếp theo, khi các bạn thấy ở câu a, sau khi đóng ngoặc biểu thức rồi . với 16 có nghĩa là nhân với 16. Tiếp, ví dụ, khi các bạn thấy trong câu g có biểu thức 2x có nghĩa là 2.x. Cuối cùng, chữ x là x chứ ko phải là dấu nhân nhé. Mik mong các bn có thể giải nhanh giúp mik bài này. Hỡi các trạng nguyên Toán, mau lại đây giải Toán nào!!!
\(a,\Rightarrow\left(35x+3\right)\cdot19=152\\ \Rightarrow35x+3=8\\ \Rightarrow x=\dfrac{1}{7}\\ b,\Rightarrow3\left(x+7\right)=42\\ \Rightarrow x+7=14\Rightarrow x=7\\ c,\Rightarrow3\left(x+1\right)=48\\ \Rightarrow x+1=16\Rightarrow x=15\\ d,\Rightarrow120-5x+100\cdot2:5=4\cdot15\\ \Rightarrow120-5x+40=60\\ \Rightarrow5x=100\Rightarrow x=20\\ e,\Rightarrow4x-10=30\\ \Rightarrow4x=40\\ \Rightarrow x=10\\ g,\Rightarrow10x+10=70\\ \Rightarrow10x=60\\ \Rightarrow x=6\)
Tìm x:
a) 4.(2-x)+x.(x+6)=x2
b) x.(x-7)-(x-2).(x+5)=0
c) (2x+3).(3-2x)+(2x-1)2=2
a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2
tìm x:
a)3(2x-3)+2(2-x)=-3
b)2x(x2-2)+x2(1-2x)-x2=-12
c)3x(2x+3)-(2x+5)(3x-2)=8
d)4x(x - 1) - 3(x2-5)-x2=(x-3)-(x+4)
e)2(3x-1)(2x+5)-6(2x-1)(x+2)=-6
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
`a,x(x-1)-(x+2)^2=1`
`<=>x^2-x-x^2-4x-4=1`
`<=>-5x=5`
`<=>x=-1`
`b,(x+5)(x-3)-(x-2)^2=-1`
`<=>x^2+2x-15-x^2+4x-4+1=0`
`<=>6x-18=0`
`<=>x-3=0`
`<=>x=3`
`c,x(2x-4)-(x-2)(2x+3)=0`
`<=>2x(x-2)-(x-2)(2x+3)=0`
`<=>(x-2)(2x-2x-3)=0`
`<=>-3(x-2)=0`
`<=>x-2=0`
`<=>x=2`
`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`
`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`
`<=>4x+26=-12`
`<=>4x=-38`
`<=>x=-19/2`
Tìm x:
a) (2x - 1) (x^2 - x + 1) = 2x^3 - 3x^2 + 2
b) (x + 1) (x^2 + 2x + 4) - x^3 - 3x^2 + 16 = 0
c) (x + 1) (x + 2) (x + 5) - x^3 - 8x^2 = 27
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
3) Tìm x của N biết a) 80 - ( 2 + 2x ) : 2 = 40 b) 11^2 - 121 = 0 c) 5^x+3 - 7 × 5^2 = 5^2 × 18 d) x^10 = x e) x^10 = 1^x g) ( 2x - 15 )^5 = ( 2x - 15 )^3 4) Thực hiện phép tính a) 5^3 ÷ 5^2 + 2^2 × 3 b) 4^3 × 125 - 125 ÷ 5^2 c) 6^2 × 28 + 72 × 6^2 d) 5^6 ÷ 5^4 + 3 × 3^2 - 80 e) 2^3 × 19 - 2^3 × 14 × 1^2021
Bài 4: tìm x:
a) \(\dfrac{4}{3}\) + (1,25 - x) = 2,25
b) \(\dfrac{17}{6}\) - (x - \(\dfrac{7}{6}\) ) = \(\dfrac{7}{4}\)
c) 4 - (2x + 1) = 3 - \(\dfrac{1}{3}\)
bài 15:
a) (\(\dfrac{-2}{3}\))9 : x = (\(\dfrac{-2}{3}\))
b) x : (\(\dfrac{4}{9}\))5 = (\(\dfrac{4}{9}\))4
c) (x + 4)3 = -125
d) (10 - 5x)3 = 64
e) (4x + 5)2 = 81
Bài 16:
a) 4 - \(1\dfrac{2}{5}\) - \(\dfrac{8}{3}\)
b) -0,6 - \(\dfrac{-4}{9}\) - \(\dfrac{16}{15}\)
c) \(-\dfrac{15}{4}\) . (\(\dfrac{-7}{15}\)) . (\(-2\dfrac{2}{5}\)
Gi ải gấp giúp mình ạ, mình rất cần gấp
Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
Tìm x:
a) (x - 5)(x + 3) = x(x - 3)
b) (x + 2)2 = (x - 1)(x + 2)
c) (x - 6)(x + 6) = x2
d) (2x - 3)2 = 4x2 - 8
a: Ta có: \(\left(x-5\right)\left(x+3\right)=x\left(x-3\right)\)
\(\Leftrightarrow x^2-2x-15-x^2+3x=0\)
\(\Leftrightarrow x=15\)
b: Ta có: \(\left(x+2\right)^2=\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow x+2=0\)
hay x=-2
c: Ta có: \(\left(x-6\right)\left(x+6\right)=x^2\)
\(\Leftrightarrow x^2-36=x^2\)(vô lý)
a. (x - 5)(x + 3) = x(x - 3)
<=> x2 + 3x - 5x - 15 = x2 - 3x
<=> x2 - x2 + 3x - 5x + 3x - 15 = 0
<=> x = 15
b. (x + 2)2 = (x - 1)(x + 2)
<=> x2 + 4x + 4 = x2 + 2x - x - 2
<=> x2 - x2 + 4x - 2x + x = -2 - 4
<=> 3x = -5
<=> \(x=\dfrac{-5}{3}\)
c. (x - 6)(x + 6) = x2
<=> x2 - 36 - x2 = 0
<=> x2 - x2 = 36
<=> 0 = 36 (vô lí)
Vậy nghiệm của PT là \(S=\varnothing\)
d. (2x - 3)2 = 4x2 - 8
<=> 4x2 - 12x + 9 - 4x2 + 8 = 0
<=> 4x2 - 4x2 - 12x = -8 - 9
<=> -12x = -17
<=> \(x=\dfrac{17}{12}\)
Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)