105-5(x-5)=95
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Sửa đề: \(\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x=105\)
Ta có: \(\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x=105\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x-105=0\)
\(\Leftrightarrow\frac{x-1}{99}-1+\frac{x-3}{97}-1+\frac{x-5}{95}-1+\frac{x-7}{93}-1+\frac{x-95}{5}-1+x-100=0\)
\(\Leftrightarrow\frac{x-100}{99}+\frac{x-100}{97}+\frac{x-100}{95}+\frac{x-100}{93}+\frac{x-100}{5}+\frac{x-100}{1}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}+\frac{1}{93}+\frac{1}{5}+1\right)=0\)
mà \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}+\frac{1}{93}+\frac{1}{5}+1\ne0\)
nên x-100=0
hay x=100
Vậy: x=100
\(\left(2x-6\right)\left(x^2+2\right)=\left(2x-6\right)\left(8x-10\right)\)
\(\Leftrightarrow\left(2x-6\right)\left(x^2+2\right)-\left(2x-6\right)\left(8x-10\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(x^2+2-8x+10\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x^2-6x-2x-12\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x-6\right)\left(x-2\right)=0\)
\(\Rightarrow x\in\left\{3;6;2\right\}\)
\(\left(5x-1\right)^2=\left(3x+5\right)^2\)
\(\Leftrightarrow\left(5x-1\right)^2-\left(3x+5\right)^2=0\)
\(\Leftrightarrow\left(5x-1-3x-5\right)\left(5x-1+3x+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(8x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\8x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{2}\end{cases}}}\)
A = 5 + 10 + 15 + ... + 95 + 100 + 105
A = 105 + 100 + 95 + ... + 15 + 10 + 5
A = (105 + 5) + (100 + 10) + (95 + 15) + ... + (50 + 60) + 55 (10 cặp)
A = 110 + 110 + 110 + ... + 110 + 55 (10 số 110)
A = 110 * 10 + 55
A = 1155
\(a.83-\left(-756-17\right)+\left(50-756\right)=83+756+17+50-756=\left(83+17\right)+50+\left(756-756\right)=100+50=150\)
\(b.-105+\left(-34-95\right)-166=-105+\left(-34\right)-95-166=\left(-105-95\right)-\left(34+166\right)=\left(-200\right)-200=-400\)
\(c,-\left(39+228-407\right)+\left(118-161\right)=-39-228+407+118-161=\left(-39-161\right)-\left(228-118\right)+407=\left(-200\right)-110=-310\)
\(d,5^{10}:5^8+60:12+\left(-10\right)=5^2+5+\left(-10\right)=5\left(5+1-2\right)=5.4=20\)
\(e,-342-\left(161-342\right)-39=-342-161+342-39=\left(-342+342\right)-\left(161+39\right)=-200\)\(g,7^5:7^3+\left(-187-149\right)-213=7^2+\left(-187\right)-149-213=\left(49-149\right)-\left(187+213\right)=\left(-100\right)-400=-500\)
a) Ta có \(\dfrac{72}{95}\times\dfrac{105}{105}=\dfrac{72}{95}\times1=\dfrac{72}{95}=\dfrac{72}{95}\)
Vậy \(\dfrac{72}{95}=\dfrac{72}{95}\times\dfrac{105}{105}\)
b) Ta có \(\dfrac{72}{95}\times\dfrac{99}{98}=\dfrac{72}{95}\times\left(1+\dfrac{1}{98}\right)=\dfrac{72}{95}\times1+\dfrac{72}{95}\times\dfrac{1}{98}=\dfrac{72}{95}+\dfrac{72}{95}\times\dfrac{1}{98}>\dfrac{72}{95}\)Vậy \(\dfrac{72}{95}< \dfrac{72}{95}\times\dfrac{99}{98}\)
a) 72/95 ... 72/95 x 105/105
72/95 ... 72/95 x 1 ( 105/105 = 1 )
72/95 = 72/95 x 105/105
b) 72/95 ... 72/95 x 99/98
72/95 x 1 ... 72/95 x 99/98 ( 99/98 > 1 )
72/95 < 72/95 x 99/98
A.6
Vì vế thứ 1 có tận cùng bằng 1 cộng với vế thứ 2 có tận cùng bằng 5
Vậy tận cùng của phép tính trên là 6
Chắc chắn đúng!!
k cho mik nha
\(95-105:x=60\)
\(105:x=95-60\)
\(105:x=35\)
\(x=105:35\)
\(x=3\)
105 - 5 ( x - 5 ) = 95
5 ( x - 5 ) = 105 - 95
5 ( x - 5 ) = 10
x - 5 = 10 : 5
x - 5 = 5
x = 5 + 5
x = 10