Bài 2 Tìm X biết
8 ) | x + 12 | - 12 = 1
9 ) 135 - | 9- x | = 35
10) 17 + x - ( 352 - 400 ) = - 32
11) 2130 - ( x + 136 ) + 72 = - 64
12) ( x - 2 ) - ( -8 ) = - 137
13) 10 - | x + 3 | = - 4 . ( - 10 )
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\(a,347\cdot2^2-2^2\cdot\left(216+184\right):8\)
\(< =>1388-4\cdot400:8\)
\(< =>1388-1600:8\)
\(< =>1388-200\)
\(< =>1188\)
\(b,132-\left[116-\left(132-128\right)^2\right]\)
\(< =>132-\left[116-\left(4^2\right)\right]\)
\(< =>132-100\)
\(< =>32\)
\(c,\left[184:\left(96-124:31\right)-2\right]\cdot3651\)
\(< =>\left[184:\left(96-4\right)-2\right]\cdot3651\)
\(< =>\left[184:92-2\right]\cdot3651\)
\(< =>\left[2-2\right]\cdot3651\)
\(< =>0\cdot3651\)
\(< =>0\)
\(e,\left(2+4+6+8+...+2020\right)\cdot\left(36\cdot333-108\cdot111\right)\)
\(< =>\left[\left(2020-2\right):2+1\right]\cdot\left(36\cdot333-108\cdot111\right)\)
\(< =>1010\cdot\left(11988-11988\right)\)
\(< =>1010\cdot0\)
\(< =>0\)
\(g,1024:2^4+140:\left(38+2^5\right)-7^{23}:7^{21}\)
\(< =>64+140:70-49\)
\(< =>64+2-49\)
\(< =>66-49\)
\(< =>17\)
\(h,\left(44\cdot52\cdot60\right):\left(11\cdot13\cdot15\right)\)
\(< =>137280:2145\)
\(< =>64\)
\(j,\left(2^{17}+15^4\right)\cdot\left(3^{19}-2^{17}\right)\cdot\left(2^4-4^2\right)\)
\(< =>\left(131072+50625\right)\cdot\left(1162261467-131072\right)\cdot\left(16-16\right)\)
\(< =>181697\cdot1162130395\cdot0\)
\(< =>0\)
mk chuc ban hoc tot nhe :))
Ta có: ^ACB + ^ACD = 180^o
=> ^ACD = 180^o - ^ ACB =180^o - 30^o =150^o
CM // AB => ^MCD = ^ABC = 75 ^o
=> ^ ACM = ^ACD - ^MCD = 150^o - 75^o = 75^o
=> ^ACM = ^MCD
=> CM là phân giác góc ACD
\(a,\frac{x-1}{x+5}=\frac{6}{7}\)
\(\Rightarrow7x-7=6x+30\)
\(\Rightarrow x=37\)
Vậy x=37
b) \(\frac{x^2}{6}=\frac{24}{25}\)
\(\Leftrightarrow x^2.25=6.24\)
\(\Leftrightarrow x^2.25=144\)
\(\Leftrightarrow x^2=\frac{144}{25}\)
\(\Leftrightarrow x=\pm\sqrt{\frac{144}{25}}=\pm\frac{12}{5}\)
Vì x>2 suy ra 2-x<0
Suy ra |2-x|=x-2
Vì x>2 suy ra x+1>3
Suy ra |x+1|=x+1
B=-x-1+x-2=-3
Vì x > 2 => 2 - x < 0
=> 2 - x = x - 2
Vì x > 2 => x + 1 > 3
=> x+1= x+1
B = -x - 1 + x - 2 = -3
hok tốt
*YOUTUBER*
=> 2x :(-0,35)=-21/20
=> 2x=0,3675
=> x=0,18375
study wel
k nha
mk xin cảm ơn mọi ngời nhiều
\(2x:\left(-0,35\right)=-1,4:1\frac{1}{3}\)
\(\Rightarrow2x:-\frac{7}{20}=-\frac{7}{5}:\frac{4}{3}\)
\(\Rightarrow2x:-\frac{7}{20}=-\frac{21}{20}\)
\(\Rightarrow2x=-\frac{21}{20}.\left(-\frac{7}{20}\right)\)
đến đêy thì gặp vấn đề ^^"
\(C\left(x\right)=\frac{4x-3}{6}-\frac{5-3x}{3}+\frac{1}{3}\)
\(\frac{4x-3}{6}-\frac{5-3x}{3}+\frac{1}{3}=0\)
\(4x-3-2\left(5-3x\right)+2=0\)
\(4x-1-2\left(5-3x\right)=0\)
\(4x-1-10+6x=0\)
\(10x-11=0\)
\(10x=0+11\)
\(10x=11\)
\(x=\frac{11}{10}\)
Tim x
\(\left(-3,25\right):\left(\frac{-1}{3}x\right)=\left(-2\frac{1}{3}\right):\left(-0,25\right)\)
=> -3,25;(-1/3x)=28/3
=> -1/3x=28/3*(-3,25)
=> -1/3x=-91/3
=>x=-91/3:(-1/3)
=> x=91
vậy x=91
study well
k nha
mk xin cảm ơn mọi ng nhiều
(-3,25) : (\(-\frac{1}{3}x\)) = (\(-2\frac{1}{3}\)) : (-0,25)
<=> \(-\frac{1}{3}x\)= \(-\frac{39}{112}\)
<=>x = \(\frac{117}{112}\)
8 ) | x + 12 | - 12 = 1
\(|x+12|=13\)
\(\Rightarrow\orbr{\begin{cases}x+12=13\\x+12=-13\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-25\end{cases}}\)
Vậy \(x\in\left\{1;-25\right\}\)
9) 135 -\(|9-x|=35\)
\(|9-x|=135-35\)
\(|9-x|=100\)
\(\Rightarrow\orbr{\begin{cases}9-x=100\\9-x=-100\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-91\\x=109\end{cases}}\) Vậy x\(\in\left\{-91;109\right\}\) 10) 17+x-(352-400)=-32 17+x-352+400=-32 17+x-352 =(-32) - 400 17+x-352 =-432 17+x =(-432) + 352 17+x =-80 x =(-80) - 17 x =-97 Vậy x =-97 11) 2130 - (x+136) + 72 =-64 2130 - (x+136) =(-64) -72 2130 - (x+136) =-136 (x +136) = 2130 -(-136) x+136 =2266 x =2266 - 136 x =2130 Vậy x=2130 do sap an cơm nên chiều mình sẽ giải tiếp nha! sorryy
12) (x-2) - (-8) = -137
(x-2) +8 =-137
(x-2) =(-137) - 8
x-2 =-145
Vậy x=-145
13) 10-\(|x+3|=-4.\left(-10\right)\) 10- \(|x+3|=40\) \(|x+3|=\) 40+10 \(|x+3|=50\)
\(\Rightarrow\orbr{\begin{cases}x+3=50\\x+3=-50\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=50-3\\x=\left(-50\right)-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=20\\x=-53\end{cases}}\)
Vậy x\(\in\left\{20;-53\right\}\)