1.Tìm x
a)15-x=16
x=15-16
x=(-1)
bài 1 tính
a, 24 x 66 - 24 + 36 x 24
b, 29 x 173 - 29 x 72 -29
c, 1+5+10+15+...+55+60
d, 1x2+2x3+3x4 +...+18x19+19x20
bài 2 tìm x
a,106- ( x+ 7) =9
b, 2 x ( x+ 4) + 5 =65
c, (16x x -32) x 45=0
d, x+4 x x = 100 : 5
bài 2 tìm x
a,106- ( x+ 7) =9
x+7 = 106 - 9
x+7 = 107
x= 107 - 7
x=100
b, 2 x ( x+ 4) + 5 =65
2 x (x+4) = 65 - 5
2 x (x+4) = 60
x+4 = 60:2
x+4= 30
x= 30 - 4
x=26
c, (16x x -32) x 45=0
16 x X - 32 = 0: 45
16 x X - 32 =0
16 x X = 0 + 32
16 x X = 32
X= 32:16
X=2
d, x+4 x x = 100 : 5
X + 4 x X = 20
(1+4) x X = 20
5 x X = 20
X= 20:5
X=4
Bài 1:
a, 24 x 66 - 24 + 36 x 24
= 24 x (66-1+36)
= 24 x 101
= 24 x 100 + 24
= 2400+24
=2424
b, 29 x 173 - 29 x 72 - 29
= 29 x (173 - 72 -1)
= 29 x 100
= 2900
c, 1+5+10+15+...+55+60
= (5+60) x 6 + 1 (có 6 cặp)
= 65 x 6 + 1
= 390 +1 =391
d, D= 1 x 2 + 2 x 3 + 3 x 4 +...+18 x 19 + 19 x 20
3D= 1 x 2 x 3+ 2 x 3 x (4-1) + 3 x 4 x (5-2) +...+ 18 x 19 x (20 - 17) + 19 x 20 x (21-18)
3D= 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 +.... + 19 x 20 x 21 - 18 x 19 x 20
3D= 19 x 20 x 21 =7980
=> D= 7980:3=2660
Tìm x
1 3 × x = 1 6 x - 1 5 = 3 10
Tính f(15) biết: f(x) = x6 - 16x5 + 16x4 - 16x3 + 16x2 - 16x - 100
Tìm GTLN của \(\frac{16x^2-1}{x^2+15}\)
Biểu thức này không có max bạn nhé. Bạn xem lại đề.
tìm x biết 16x^2-(4x-5)^2=15
\(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow16x^2-\left(16x^2-40x+25\right)=15\)
\(\Leftrightarrow16x^2-16x^2+40x-25=15\)
\(\Leftrightarrow40x=40\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
16x2 - ( 4x - 5 )2 = 15
<=> 16x2 - ( 16x2 - 40x + 25 ) = 15
<=> 16x2 - 16x2 + 40x - 25 = 15
<=> 40x - 25 = 15
<=> 40x = 40
<=> x = 1
<=> x =
Tìm x :
a) ( 2x + 3 )^2 - 4( x - 1 )( x + 1 ) = 49
b) 16x^2 - ( 4x - 5 )^2 = 15
c) ( 2x + 1 )^2 - ( x - 1)^2 = 0
a) ( 2x + 3 )^2 - 4( x - 1 )( x + 1 ) = 49
=>4x2+12x+9-4x2+4=49
=>12x+13=49
=>12x=36
=>x=3
b) 16x^2 - ( 4x - 5 )^2 = 15
=>16x2-16x2+40x-25=15
=>40x-25=15
=>40x=40
=>x=1
c) ( 2x + 1 )^2 - ( x - 1)^2 = 0
=>4x2+4x+1-x2+2x-1=0
=>3x2+6x=0
=>3x(x+2)=0
=>3x=0 hoặc x+2=0
=>x=0 hoặc x=-2
a) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ =>4x^2+12x+9-4x^2+4=49\\=>12x+13=49\\ =>12x=36\\ =>x=3\)
b) \(16x^2-\left(4x-5\right)^2=15\\ =>16x^2-16x^2+40x-25=15\\ =>40x-25=15\\ =>40x=40\\ =>x=1\)
c) \(\left(2x+1\right)^2-\left(x-1\right)^2=0\\ =>4x^2+4x+1-x^2+2x-1=0\\ =>3x^2+6x=0\\ =>3x\left(x+2\right)=0\\ =>\left[\frac{3x=0}{x+2=0}\right]=>\left[\frac{x=0}{x=-2}\right]\)
Tìm x biết,
2|x+5|+x=22
x^2+16x+15=0
x^2+16x+15=0
=>x^2+x+15x+15=0
=>(x^2+x)+(15x+15)=0
=>x(x+1)+15(x+1)=0
=>(x+1)(x+15)=0
=>x+1=0 hoặc x+15=0
=>x=-1 hoặc x=-15
2|x+5|+x=22
=>x=-32 hoặc 2
x^2+16x+15=0
C1:đl vi-ét
tổng các nghiệm
x1+x2\(=-\frac{b}{a}=-16\)
tích các nghiệm
x1+x2=\(\frac{c}{a}=15\)
=>x=-15;-1
C2:denta
<=>162-4(1*15)=196
\(x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-16\pm\sqrt{196}}{2}\)
=>x=-15;-1
2/x+5/+x=22==>x=a hoặc x=-a với x=a ta có:2(a+5)+a=22==>a=4.với x=-a ta có:2(-a-5)--a=22==>a=-32......................................................
Giải phương trình:
1, \(\frac{x^2-13x+15}{x^2-14x+15}+\frac{x^2-15x+15}{x^2-16x+16}=\frac{1}{12}\)
tìm x
17x – ( -16x – 37) = 2x +
-2x –3. (x – 17) = 34 – 2(-x + 25
17x + 3. ( -16x – 37) = 2x + 43 - 4x
103 -57: [-2. (2x – 1)2 – (-9)0] = -106
3x – 32 > -5x + 1
15 + 4x < 2x – 145
-3. (2x + 5) -16 < -4. (3 – 2x)
-2x + 15 < 3x – 7 < 19 – x
x + (x+1) + (x+2) + (x+3) + .... + 13 + 14 = 14
25 + 24 + 23 +...+ x + (x - 2) + (x – 3) = 25
17x + 3. ( -16x – 37) = 2x + 43 - 4x
<=>17x-48x-111=-2x+43
<=>-29x=154
<=> \(x=-\frac{154}{29}\)
-3. (2x + 5) -16 < -4. (3 – 2x)
\(\Leftrightarrow-6x-31< -12+8x.\)
\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)
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hỏi google là đc hết mak
1) x2 - 11x + 3
2) 1+7x3
3) x3 + 3x2 - 16x - 48
4) x3 - x2 – x - 1
5) x3 + 2x2 - 2x - 1
6) 4x(x - 3y )+ 12y(3y - x)
3: \(x^3+3x^2-16x-48\)
\(=x^2\left(x+3\right)-16\left(x+3\right)\)
\(=\left(x+3\right)\left(x-4\right)\left(x+4\right)\)