1.53x201=10653

2.

a)4 xX-15x4=6x15+X

<=>4x-60=90+x

<=>4x-x=60+90

<=>3x=150

<=>x=50

b(102+28-2x):20-5=1

<=>(130-2x):20=6

<=>130-2x=120

<=>2x=10

<=>x=5

c)1999x-x=1999+1997+1999

<=>1998x=5995

<=>\(x\approx3\)

d)105-5x=150:15

<=>105-5x=10

<=>5x=95

<=>x=19

e)(x+1)+(x+2)+(x+3)+...+(x+6)=78

<=>6x+21=78

<=>6x=57

<=>x=9,5

1) 53x201=53x200+53=10600+53=10653

2a)

4x-15x4=6x15+x

4x-60=90+x

4x-x=90+60

3x=150

x=50

b) x=5

c) x=3

d)x=19

e) x=9,5

các bạn ơi 4x và 3x là j vậy ??

#maianhhappy

bài 1 tính giá trị biểu thức

( - 25 ) nhân ( -3 ) nhân x với x = 4

\(\left(-25\right).\left(-3\right).4\)

\(=\left(-25\right).4.\left(-3\right)\)

\(=-100.\left(-3\right)=300\)

( -1 ) nhân ( -4 ) nhân 5 nhân 8 nhân y với y =25

\(\left(-1\right).\left(-4\right).5.8.25\)

\(=4.5.8.25=4.25.5.8\)

\(=100.40=40000\)

( 2ab mũ 2 ) : c với a =4 ; b= -6 ; c =12

\(\left(2.4.\left(-6\right)\right)^2:12\)

\(=\left(-48\right)^2:12\)

\(=2304:12=192\)

[ ( -25 ) nhân ( - 27 ) nhân ( -x ) ] : y với x = 4 ; y = -9

\(\left[\left(-25\right).\left(-27\right).\left(-4\right)\right]:-9\)

\(=-2700:\left(-9\right)\)

\(=300\)

(a mũ 2 _ b mũ 2) : ( a + b ) nhân ( a _ b ) với a + 5 , b = -3

\(\left(5^2-\left(-3\right)^2\right):\left(5-3\right).\left(5+3\right)\)

\(=16:2.8\)

\(=8.8=64\)

bài 2 tìm x

( 2x _ 5 ) + 17 = 6

\(2x-5=-11\)

\(2x=-6\)

\(x=-3\)

10 _ 2 ( 4 _ 3x ) = - 4

\(2.\left(4-3x\right)=14\)

\(4-3x=7\)

\(3x=-3\)

\(x=-1\)

- 12 + 3 ( -x + 7 ) = -18

\(3\left(-x+7\right)=-6\)

\(-x+7=-2\)

\(-x=-9\)

\(x=9\)

24 : ( 3x _ 2 ) = -3

\(3x-2=-8\)

\(3x=-6\)

\(x=-2\)

- 45 : 5 nhân ( -3 _ 2x ) = 3

\(5.\left(-3-2x\right)=-15\)

\(-3-2x=-3\)

\(2x=0\)

\(x=0\)

ai nhNH MK K

...

;;;;;

'''''

...

sory tớ bị đau tay không viết được nhiều !

a) ĐKXĐ : \(3\le x\le7\)

Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)

\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)

Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)

\(1,\\ a,A\le\sqrt{\left(x-3+7-x\right)\left(1+1\right)}=\sqrt{8}=2\sqrt{2}\\ A^2=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4\Leftrightarrow A\ge2\\ \Leftrightarrow2\le A\le2\sqrt{2}\\ \left\{{}\begin{matrix}A_{min}\Leftrightarrow\left(x-3\right)\left(7-x\right)=0\Leftrightarrow...\\A_{max}\Leftrightarrow x-3=7-x\Leftrightarrow x=5\end{matrix}\right.\)

\(B=\dfrac{\dfrac{5}{2}\left(4x^4+4x^2+1\right)+2\left(x^4-x^2+\dfrac{1}{4}\right)}{\left(2x^2+1\right)^2}\\ B=\dfrac{\dfrac{5}{2}\left(2x^2+1\right)^2+2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}=\dfrac{5}{2}+\dfrac{2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}\ge\dfrac{5}{2}\)

\(B=\dfrac{3\left(4x^4+4x^2+1\right)-4x^2}{\left(1+2x^2\right)^2}=\dfrac{3\left(1+2x^2\right)^2-4x^2}{\left(1+2x^2\right)^2}=3-\dfrac{4x^2}{\left(1+2x^2\right)^2}\)

Vì \(-\dfrac{4x^2}{\left(1+2x^2\right)^2}\le0\Leftrightarrow B\le3\)

\(\Leftrightarrow\left\{{}\begin{matrix}B_{min}\Leftrightarrow x^2=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{1}{\sqrt{2}}\\B_{max}\Leftrightarrow x=0\end{matrix}\right.\)

\(2,\)

Ta có \(\left(y-2x\right)^2=\left(-2x+y\right)^2=\left[\dfrac{1}{3}\left(-6x\right)+\dfrac{1}{4}\left(4y\right)\right]^2\)

\(\Leftrightarrow\left(y-2x\right)^2\le\left[\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2\right]\left[\left(-6x\right)^2+\left(4y\right)^2\right]=\dfrac{5^2}{3^2\cdot4^2}\left(36x^2+16y^2\right)=\dfrac{5^2}{4^2}\\ \Leftrightarrow\left|y-2x\right|\le\dfrac{5}{4}\\ \Leftrightarrow-\dfrac{5}{4}\le y-2x\le\dfrac{5}{4}\\ \Leftrightarrow\dfrac{15}{4}\le y-2x+5\le\dfrac{25}{4}\)

\(Max\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{9}{20}\end{matrix}\right.\\ Min\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{9}{20}\end{matrix}\right.\)

khỏi chép đề nha

a\Câu 1:

200 : ( 15 - x ) = 15 + 5

200 : ( 15 - x ) = 20

15 - x = 200 : 20

15 - x = 10

x = 15 - 10

x = 5

Câu 2:

63 : ( x - 5 ) = 22 -1

63 : ( x - 5 ) = 21

x - 5 = 63 : 21

x - 5 = 3

x = 3 + 5

x = 8

\(\frac{10}{3}.x+\frac{67}{4}=\frac{53}{4}\)

\(\frac{10}{3}.x=\frac{53}{4}-\frac{67}{4}\)

\(\frac{10}{3}.x=-\frac{7}{2}\)

\(x=-\frac{7}{2}:\frac{10}{3}\)

\(x=-\frac{21}{20}\)