Tìm x:
c) 7\(^{x+1}+7^{x}=8\times7^5\)
d) \(11^{x+3}+11^{x+2}=12\times11^{10}\)
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vì số nào nhân vs o đều bàng 0 nên x trong cả hai bài trên đều bằng 0
1, x=0
2, x=0
Bài làm
m) (x + 2).(3 - x) = 0;
=> x + 2 = 0 hoặc 3 - x = 0
=> x = -2 hoặc x = 3
Vậy x = -2 hoặc x = 3
d) 511.712 + 511.711
= 511 . ( 712 + 711 )
= 511 . [ 711 . ( 7 + 1 ) ]
= 511 . 711 . 8
= ( 5 . 7 )11 . 8
= 3511 . 8
512.712 + 9.511.711
= 511 ( 5 . 712 + 9 . 1 . 711 )
= 511 [ 711 ( 5 . 7 + 9 . 1 . 1 ) ]
= 511 ( 711 . 44 )
= 511 . 711 . 44
= 3511 . 44
m. \(\left(x+2\right)\left(3-x\right)=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\3-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
d. \(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.\left(7^{12}+7^{11}\right)}{5^{11}.\left(5.7^{12}+9.7^{11}\right)}=\frac{7^{12}+7^{11}}{5.7^{12}+9.7^{11}}=\frac{1}{5.9}=\frac{1}{45}\)
q. \(\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10+11=11\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10=0\)
\(\Rightarrow\left[\left(x-3\right)+\left(x-2\right)+\left(x-1\right)\right]+(1+2+3+...+10)=0\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+55=0\)
\(\Rightarrow x-3+x-2+x-1=-55\)
\(\Rightarrow3x-6=-55\)
\(\Rightarrow3x=-49\)
\(\Rightarrow x=-\frac{49}{3}\)
\(\dfrac{10}{11}:\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\right)\)
\(=\dfrac{10}{11}:\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)
\(=\dfrac{10}{11}:\left(\dfrac{1}{3}-\dfrac{1}{11}\right)\)
\(=\dfrac{10}{11}:\dfrac{8}{33}\)
\(=\dfrac{10}{11}\times\dfrac{33}{8}\)
\(=5\times\dfrac{3}{4}\)
\(=\dfrac{15}{4}\)
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
a ) \(-5\times\left(-x+7\right)-3\times\left(-x-5\right)=-4\times\left(12-x\right)+48\)
\(\Leftrightarrow5x-35+3x+15=-48+4x+48\)
\(\Leftrightarrow5x-3x+4x=35-15-48+48\)
\(\Leftrightarrow2x=20\)
\(\Leftrightarrow x=10\)
b ) \(-2\times\left(15-3x\right)-4\times\left(-7x+8\right)=-5-9\times\left(-2x+1\right)\)
\(\Leftrightarrow-30+6x-28x-32=-5+18x-9\)
\(\Leftrightarrow6x-28x-18x=30+32-5-9\)
\(\Leftrightarrow-40x=48\)
\(\Leftrightarrow x=-1.2\)
Bài 1:
a. $(-20)+x=-30$
$x-20=-30$
$x=-30+20=-(30-20)=-10$
b.
$(-10)-x=-20$
$x=(-10)-(-20)=-10+20=20-10=10$
c. Đề sai. Bạn xem lại.
d.
$x+(-3)=-7$
$x=-7-(-3)=-7+3=-(7-3)=-4$
e.
$x-(-5)=-9$
$x=(-9)+(-5)=-14$
f.
$x(-11)=12$
$x=\frac{12}{-11}=\frac{-12}{11}$
h.
$2x-10=20$
$2x=20+10=30$
$x=30:2=15$
l.
$4x-8=-8$
$4x=-8+8=0$
$x=0:4=0$
k.
$-12-(-2)x=-8$
$(-2)x=-12-(-8)=-12+8=-(12-8)=-4$
$x=(-4):(-2)=2$
Bài 2:
a. $-20-(10-x)=-3$
$10-x=-20-(-3)=-20+3=-(20-3)=-17$
$x=10-(-17)=10+17=27$
b.
$14+(14-x)=-2$
$14-x=-2-14=-16$
$x=14-(-16)=14+16=30$
c.
$-15-(x-3)=-7$
$x-3=-15-(-7)=-15+7=-8$
x=-8+3=-5$
d.
$(x+4)+(-20)=-8$
$x+4=-8-(-20)=-8+20=12$
$x=12-4=8$
e.
$-2x-2=-4$
$-2x=-4+2=-2$
$x=(-2):(-2)=1$
f.
$-2x+4=-4$
$-2x=-4-4=-8$
$x=(-8):(-2)=4$
l.
$-12-(-2)x=-2-4=-6$
$(-2)x=-12-(-6)=-12+6=-6$
$x=(-6):(-2)=3$
\(a)\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{-11.13^7}{11^5.13^8}=\frac{-1}{11^4.13}\) (Bạn xem thử xem có sai đề không nhé)
\(c)\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3+1\right)}{2^9.3^{10}}=\frac{2.4}{3}=\frac{8}{3}\)
\(d)\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.4+9\right)}=\frac{8}{20+9}=\frac{8}{29}\)
\(a)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{\left(-11\right)\cdot13^7}{11^5\cdot13^8}=\frac{-1}{11^4\cdot13}=\frac{-1}{14641\cdot13}=\frac{-1}{190333}\)
\(c)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\cdot2}{2^9\cdot3^{10}}=\frac{2\cdot2}{3}=\frac{4}{3}\)
Câu c)
\(7^{x + 1} + 7^{x} = 8 \times 7^{5}\)
Bước 1: Đặt \(7^{x} = a\)
\(7^{x + 1} = 7 \cdot 7^{x} = 7 a\)
Bước 2: Thay vào phương trình
\(7 a + a = 8 \times 7^{5}\) \(8 a = 8 \times 7^{5}\)
Bước 3: Chia cả hai vế cho 8
\(a = 7^{5}\)
Bước 4: Trả lại \(a = 7^{x}\)
\(7^{x} = 7^{5} \Rightarrow x = 5\)
✅ Kết quả: \(x = 5\)
Câu d)
\(11^{x + 3} + 11^{x + 2} = 12 \times 11^{10}\)
Bước 1: Đặt \(11^{x} = b\)
\(11^{x + 3} = 11^{3} \cdot 11^{x} = 1331 b\) \(11^{x + 2} = 11^{2} \cdot 11^{x} = 121 b\)
Bước 2: Thay vào phương trình
\(1331 b + 121 b = 12 \times 11^{10}\) \(1452 b = 12 \times 11^{10}\)
Bước 3: Chia cả hai vế cho 1452
Trước hết:
\(1452 = 12 \times 121\)
nên:
\(b = \frac{12 \times 11^{10}}{12 \times 121} = \frac{11^{10}}{11^{2}} = 11^{8}\)
Bước 4: Trả lại \(b = 11^{x}\)
\(11^{x} = 11^{8} \Rightarrow x = 8\)
✅ Kết quả: \(x = 8\)
c: \(7^{x+1}+7^{x}=8\cdot7^5\)
=>\(7^{x}\cdot7+7^{x}=8\cdot7^5\)
=>\(8\cdot7^{x}=8\cdot7^5\)
=>\(7^{x}=7^5\)
=>x=5
d: \(11^{x+2}+11^{x+3}=12\cdot11^{10}\)
=>\(11^{x+2}+11^{x+2}\cdot11=12\cdot11^{10}\)
=>\(11^{x+2}\cdot12=11^{10}\cdot12\)
=>\(11^{x+2}=11^{10}\)
=>x+2=10
=>x=8