Tìm x, biết 1 / 4 x = 16
A. x = -2 B. x = 2
C. x = 1/2 D. x = -1/2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TQ
11 tháng 6 2021
a) \(\sqrt{x}< 3\)<=> x<9
b)\(\sqrt{4-x}\) ≤ 2 <=> 4 - x ≤ 4 <=> x≥0
c)\(\sqrt{x+2}=\sqrt{4-x}\) <=> x+2=4-x <=>2x=2<=>x=1
Vậy x=1
d)\(\sqrt{x^2-1}\)=x-1 <=> x\(^2\)-1=x\(^2\)-2x+1 <=> x\(^2\)-\(x^2\)-2x+1+1=0 <=> 2x=2 <=> x=1
Vậy x=1
NN
16 tháng 7 2019
a. 16a2 - 49.( b - c )2
= ( 4a )2 - 72.( b - c )2
= ( 4a )2 - [ 7.( b - c ) ]2
= ( 4a )2 - ( 7b - 7c )2
= ( 4a - 7b + 7c ).( 4a + 7b - 7c )
b. ( ax + by )2 - ( ax - by )2
=( ax + by + ax - by ).( ax + by - ax + by )
= 2ax . 2by
= 2.( ax + by )
c.a6 - 1
= ( a3 )2 - 1
= ( a3 - 1 ).( a3 + 1 )
= ( a - 1 ).( a2 + a + 1 ).( a + 1 ).( a2 - a + 1 )
d. a8 - b8
= ( a4 )2 - ( b4 )2
= ( a4 - b4 ).( a4 + b4 )
= [ ( a2 )2 - ( b2 )2 ].( a4 + b4 )
= ( a2 - b2 ).( a2 + b2 ).( a4 + b4 )
= ( a - b ).( a + b ).( a2 + b2 ).( a4 + b4 )
B2
( x - 4 )2 - 36 = 0
\(\Leftrightarrow\) ( x - 4 )2 = 36
\(\Leftrightarrow\) ( x - 4 )2 = 62
\(\Leftrightarrow\) x + 4 = \(\pm\) 6
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)
Vậy x = 10 , x = -2
b. ( x - 8 )2 = 121
\(\Leftrightarrow\) ( x - 8 )2 = 112
\(\Leftrightarrow\) x - 8 = \(\pm\)11
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-8=11\\x-8=-11\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=19\\x=-3\end{cases}}\)
Vậy x = 19 , x = -3
c. x2 + 8x + 16 = 0
\(\Leftrightarrow\)x2 + 2.4x + 42 = 0
\(\Leftrightarrow\) ( x + 4 )2 = 0
\(\Leftrightarrow\) x + 4 = 0
\(\Leftrightarrow\) x = -4
Vậy x = -4
d. 4x2 - 12x = - 9
\(\Leftrightarrow\)( 2x )2 - 2.2.x.3 + 32 = 0
\(\Leftrightarrow\) ( 2x - 3 )2 = 0
\(\Leftrightarrow\) 2x - 3 = 0
\(\Leftrightarrow\) 2x = 3
\(\Leftrightarrow\) \(x=\frac{3}{2}\)
Vậy x = \(\frac{3}{2}\)
2 tháng 10 2016
a) \(4a^3b^3c^2x+12a^3b^4c^2-16a^4b^5cx\)
\(=4a^3b^3c\left(cx+3bc-4ab^2x\right)\)
b) \(\left(b-2c\right)\left(a-b\right)-\left(a+b\right)\left(2c-b\right)\)
\(=\left(b-2c\right)\left(a-b+a+b\right)=2a\left(b-2c\right)\)
c) \(3a\left(a+5\right)-2\left(5+a\right)=\left(a+5\right)\left(3a-2\right)\)
d) \(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)\)
2 tháng 11 2021
\(b,=\left(x+8-x+2\right)^2=100\\ c,=x^2\left(x^2-16\right)-x^4+1=x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)
2 tháng 11 2021
b) \(=\left(x+8-x+2\right)^2=10^2=100\)
c) \(=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\)
d) \(=x^3+1-x^3+1=2\)
28 tháng 5 2022
a: \(=4x^2+20x+25+4x^2-20x+25-\left(4x^2-1\right)\)
\(=8x^2+50-4x^2+1=4x^2+51\)
b: \(=8a^3+12a^2b+6ab^2+b^3+8a^3-12a^2b+6ab^2-b^3-16a^3\)
\(=12ab^2\)
c: \(\left(2x-1\right)^3-\left(x-2\right)\left(x^2+2x+4\right)-7x^3-2x\)
\(=\left(2x-1\right)^3-x^3+8-7x^3-2x\)
\(=8x^3-12x^2+6x-1-8x^3-2x+8\)
\(=-12x^2+4x+7\)
d: \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+3\)
\(=-3x^2+4x+3\)
24 tháng 8 2021
Bài 3:
a) \(4x^2+4x+1=\left(2x+1\right)^2\)
b) \(9x^2-12x+4=\left(3x-2\right)^2\)
c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)
27 tháng 10 2021
Bài 1:
\(a,=15x^4-12x^3+9x^2\\ b,=-15x^3y^2+25x^2y^2-5xy^3\\ c,=5x^3-15x^2-4x^2+12x=5x^3-19x^2+12x\\ d,=3x^3-9x^2y+xy^2-3y^3+5x^2y-15xy^2=3x^3-3y^3-4x^2y-14xy^2\)
Bài 2:
\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=x^2+16x+64-2x^2-12x+32+x^2-4x+4=100\\ c,=x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)
9 tháng 3 2022
Với x >= 0 ; x khác 16
\(B=\dfrac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\dfrac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+1}\)
15 tháng 3 2021
Bài 1:
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)
\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)
Suy ra: \(-12x-3=8x-2-6x-8\)
\(\Leftrightarrow-12x-3-2x+10=0\)
\(\Leftrightarrow-14x+7=0\)
\(\Leftrightarrow-14x=-7\)
\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
Đáp án A