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a) \(435-\left(x+16\right)=425:17\)
\(435-\left(x+16\right)=25\)
\(x+16=435-25\)
\(x+16=410\)
\(x=410-16\)
\(x=394\)
b) \(\left(x+\frac{3}{4}\right)\times\frac{7}{4}+\frac{7}{6}=5\)
\(\left(x+\frac{3}{4}\right)\times\frac{7}{4}=5-\frac{7}{6}\)
\(\left(x+\frac{3}{4}\right)\times\frac{7}{4}=\frac{23}{6}\)
\(x+\frac{3}{4}=\frac{23}{6}:\frac{7}{4}\)
\(x+\frac{3}{4}=\frac{46}{21}\)
\(x=\frac{46}{21}-\frac{3}{4}\)
\(x=\frac{121}{84}\)
435 - ( x + 16 ) = 425 : 17
435 - ( x + 16 ) = 25
x + 16 = 435 - 25
x + 16 = 410
x = 410 - 16
x = 394
a) Số các số hạng :
[(x + 16) - (x + 1)] + 1 = 16 (số hạng)
Tổng trên là :
[(x + 16) + (x + 1)] x 16 : 2 = (2x + 17) x 16 : 2 = 154
=> (2x + 17) x 16 = 208
=> 2x + 7 = 13
=> 2x = 20
=> x = 10
a) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + .... + ( x + 16 ) = 154
=> x + 1 + x + 2 + x + 3 + ... + x + 16 = 154
=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 16 ) = 154
=> 16x + 136 = 154
=> x = ( 154 - 136 ) : 16
=> x = 9/8
b) ( x + 1 ) + ( x + 3 ) + ( x + 5 ) + ... + ( x + 19 ) = 245
=> x + 1 + x + 3 + x + 5 + ... + x + 19 = 245
=> ( x + x + x + ... + x ) + ( 1 + 3 + 5 + ... + 19 ) = 245
=> 10x + 100 = 245
=> x = ( 245 - 100 ) : 10
=> x = 9/2
a) \(435-\left(x+16\right)=425:17\)
\(435-\left(x+16\right)=25\)
\(\Rightarrow x+16=435-25=410\)
\(\Rightarrow x=410-16=394\)
b) \(\left(x+\frac{4}{3}\right)\cdot\frac{7}{4}=5-\frac{7}{6}\)
\(\left(x+\frac{4}{3}\right)\cdot\frac{7}{4}=\frac{23}{6}\)
\(\Rightarrow x+\frac{4}{3}=\frac{23}{6}:\frac{7}{4}=\frac{46}{21}\)
\(\Rightarrow x=\frac{46}{21}-\frac{4}{3}=\frac{6}{7}\)
Ủng hộ mk nka!!!^_^^_^^_^
a) 435 - ( x + 16 ) = 25
x+ 16 = 435 - 25
x + 16 = 410
x = 410 - 16
x = 394
b) (x+4/3 ) * 7/4 = 23/6
x+4/3= 23/6 : 7/4
x+4/3 = 46/21
x = 46/21 - 4/3
x = 6/7
17- {-x-[-x-(-x)]}= -16
17- {-x-0}=-16
17+x=-16
x=-16-17
x=-32
Vậy x=-32
tick cho mình tròn 510 nha
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
Bài 2:
a, \(\dfrac{5}{23}\) \(\times\) \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) \(\times\) \(\dfrac{9}{26}\)
= \(\dfrac{5}{23}\) \(\times\) ( \(\dfrac{17}{26}\) + \(\dfrac{9}{26}\))
= \(\dfrac{5}{23}\) \(\times\) \(\dfrac{26}{26}\)
= \(\dfrac{5}{23}\)
b, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)
\(\frac{21}{11}\)x \(\frac{22}{17}\)x\(\frac{68}{63}\)
= \(\frac{42}{17}\)x\(\frac{68}{63}\)
= \(\frac{8}{3}\)
18 - x : 2 = 16
x : 2 = 18 - 16
x = 2 x 2
x = 4
\(\frac{21}{11}\times\frac{22}{17}\times\frac{68}{63}\)
\(=\frac{21}{11}\times\frac{22\times68}{17\times63}\)
\(=\frac{21}{11}\times\frac{22\times4}{1\times63}\)
\(=\frac{21}{11}\times\frac{88}{63}\)
\(=\frac{21\times88}{11\times63}\)
\(=\frac{1\times8}{1\times3}\)
\(=\frac{8}{3}\)
\(18-x:2=16\)
\(x:2=18-16\)
\(x:2=2\)
\(x=2\times2\)
\(x=4\)
16 : (0,3 * x +2,5) : x = 17 + 3= 20
-> \(\frac{0,3\times x+2,5}{x}=\frac{16}{20}=\frac{4}{5}\)
\(\frac{0,3\times x}{x}+\frac{2,5}{x}=\frac{4}{5}\)
\(0,3+\frac{2,5}{x}=\frac{4}{5}\)
\(\frac{2,5}{x}=\frac{4}{5}-0,3\)
\(\frac{2,5}{x}=\frac{1}{2}\)
\(x=2,5:\frac{1}{2}=5\)
Mình thay x là a nha:
16 : (0,3 x a + 2,5) : a - 3 = 17
(16 : 0,3 x 16 :a + 26 : 2,5) : a = 17 + 3
(\(\frac{160}{3}\) x \(\frac{16}{a}\) + \(\frac{52}{5}\)) : a = 20
\(\frac{160}{3}\) : a x \(\frac{16}{a}\) :a + \(\frac{52}{5}:a\)= 20
\(\frac{160}{3a}\times\frac{16}{a\times a}+\frac{52}{5a}=20\)
...
\(17\cdot\left(x+2\right)=16\cdot\left(x+5\right)\)
\(17x+34=16x+80\)
\(x=80-34=46\)
Vậy \(x=46\)