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1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
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a) => y+42+2y= -12-14+2y
y+2y-2y = -12-14-42
y= -68
b) => 15+y-5-5y= -12-5y
y-5y+5y= -12-15+5
y = -22
c) => 2y+5-8y+21= -3-5y-2
2y-8y+5y= -3-2-5-21
-y= -31=>y=31
d)=> -13+3y+23= -120+y
3y-y= -120+13-23
2y= -130=>y= -65
e) => -21+32+5y= 16+4y
5y-4y= 16+21-32
y= 5
bài 1
a)y-(-42-2y) = (-12) - 14 +2y
y +42 + 2y = -12 -14 +2y
3y + 42 = -26 +2y
y = -68
b)15-(-y+5)-5y=-(12+5y+2)
15+y-5-5y=-12-5y-2
10-4y=-14-5y
-4y+5y=-14-10=-24
c)2y-(-5+8y-21)=-3-(5y+2)
2y+5-8y+21=-3y-5y-2
-6y+26=-8y-2
-6y+8y=-2-26
2y=-28
y=-28/2=-14
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Bài 2:
a: =6(15-5)=6*10=60
b: =9(-3+23)=9*20=180
c: =11(-10+210)=11*200=2200
d: =125*4+125*4=125*8=1000
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các bạn giải giùm mình nha ,nếu giúp được thì các bạn sẽ giải hết bài cho mình nha.bạn nào làm được hết mình sẽ tích đúng cho luôn.mình cần bài này trong 3 tiếng nữa .
\(x=\left(\frac{33}{5}:6-0,125\cdot8+\frac{32}{15}\cdot0,03\right)\cdot\frac{11}{4}\)
\(x=\left(\frac{33}{5}\cdot\frac{1}{6}-\frac{1}{8}\cdot8+\frac{32}{15}\cdot\frac{3}{100}\right)\cdot\frac{11}{4}\)
\(x=\left(\frac{11}{10}-1+\frac{8}{125}\right)\cdot\frac{11}{4}\)
\(x=\frac{41}{250}\cdot\frac{11}{4}\)
\(x=\frac{451}{100}\)
\(x=\left(\frac{33}{5}:6-0,125.8+\frac{32}{15}.0,03\right).\frac{11}{4}\)
\(< =>x=\left(\frac{33}{30}-1+\frac{32}{15}.\frac{3}{100}\right).\frac{11}{4}\)
\(< =>x=\left(\frac{3}{30}+\frac{32}{500}\right).\frac{11}{4}\)
\(< =>x=\left(\frac{1}{10}+\frac{8}{125}\right).\frac{11}{4}\)
\(< =>x=\frac{41}{250}.\frac{11}{4}=\frac{451}{100}\)