Odpowiedź:
zad. 1
a)
X + 3K = L |-3K
X = L - 3K
E + X = 9F |- E
X = 9F - E
4G + X = H |-4G
X = H - 4G
3P + X = 8R |-3P
X = 8R - 3P
b)
X - 7C = D |+7C
X = D + 7C
X - 5M = N |+5M
X = N + 5M
X - 3P = R |+3P
X = R + 3P
X - 4A = 9B |+4A
X = 9B + 4A
c)
7MX = N |:7M
[tex]X = \frac{N}{7M}[/tex]
3PX = R |:3P
[tex]X=\frac{R}{3P}[/tex]
PX = 2R |:P
[tex]X = \frac{2R}{P}[/tex]
3KX = 2L |:3K
[tex]X = \frac{2L}{3K}[/tex]
d)
[tex]\frac{X}{A}=\frac{1}{2}B[/tex] |*A
[tex]X = \frac{AB}{2}[/tex]
[tex]\frac{X}{A}=6B[/tex] |*A
X = 6BA
[tex]\frac{X}{5T} =U[/tex] |*5T
X = 5UT
[tex]\frac{X}{2M}=5N[/tex] |*2M
X = 5N * 2M
X = 10 MN
zad. 2
a)
5a + M = N |-M
5a = N - M |:5
[tex]a = \frac{N-M}{5}[/tex]
b)
aX - Y = Z |+Y
aX = Z + Y |:X
[tex]a = \frac{Z+Y}{X}[/tex]
c)
U + 4a = W |-U
4a = W - U |:4
[tex]a = \frac{W-U}{4}[/tex]
d)
Ba - C = D |+C
Ba = D + C |:b
[tex]a=\frac{D+C}{B}[/tex]
e)
5p + ar = 3 |-5p
ar = 3 - 5p |:r
[tex]a = \frac{3*5p}{r}[/tex]
f)
mn - 6a = b |-mn
-6a = b - mn |:(-6)
[tex]a=\frac{b-mn}{-6}[/tex]
g)
-3r + 2a = 1 |+3r
2a = 1 + 3r \:2
[tex]a = \frac{1+3r}{2}[/tex]
zad. 3
a) a + b = 2c
wyznaczamy a:
a + b = 2c |-b
a = 2c - b
wyznaczamy b:
a + b = 2c |-a
b = 2c - a
wyznaczamy c:
a + b = 2c |:2
[tex]c = \frac{a+b}{2}[/tex]
zaznacz: a = 2c - b oraz b = 2c - a
b) mn = 2p
wyznaczamy m:
mn = 2p|:n
[tex]m = \frac{2p}{n}[/tex]
wyznaczamy n:
mn = 2p |:m
[tex]n = \frac{2p}{m}[/tex]
wyznaczamy p:
mn = 2p |:2
[tex]p=\frac{mn}{2}[/tex]
zaznacz: [tex]m=\frac{2p}{n}[/tex] / [tex]p=\frac{mn}{2}[/tex] oraz [tex]n=\frac{2p}{m}[/tex]
Szczegółowe wyjaśnienie:
