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`@` `\text {Ans}`
`\downarrow`
`a)`
`(2x - 1)^2 + 1 = 26`
`\Rightarrow (2x - 1)^2 = 26 - 1`
`\Rightarrow (2x - 1)^2 = 25`
`\Rightarrow (2x - 1)^2 = (+-5)^2`
`\Rightarrow`\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=6\div2\\x=-4\div2\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-2;3\right\}\)
`b)`
`(2x - 4)^3 + 2 = 66`
`\Rightarrow (2x - 4)^3 = 66 - 2`
`\Rightarrow (2x - 4)^3 = 64`
`\Rightarrow (2x - 4)^3 = 4^3`
`\Rightarrow 2x - 4 = 4`
`\Rightarrow 2x = 8`
`\Rightarrow x = 8 \div 2`
`\Rightarrow x = 3`
Vậy, `x = 3`
`c)`
\(7^{x+2}+5\cdot7^{x+1}+15=603\)
`\Rightarrow 7^x . 7^2 + 5 . 7^x . 7 = 603 - 15`
`\Rightarrow 7^x . 7^2 + 35 . 7^x = 588`
`\Rightarrow 7^x . (7^2 + 35) = 588`
`\Rightarrow 7^x . 84 = 588`
`\Rightarrow 7^x = 588 \div 84`
`\Rightarrow 7^x = 7`
`\Rightarrow 7^x = 7^1`
`\Rightarrow x = 1`
Vậy, `x = 1.`
\(#48Cd\)
1a) (2x - 6)(x + 2) = 0
=> \(\orbr{\begin{cases}2x-6=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x=6\\x=-2\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
b) (x2 + 7)(x2 - 25) = 0
=> \(\orbr{\begin{cases}x^2+7=0\\x^2-25=0\end{cases}}\)
=> \(\orbr{\begin{cases}x^2=-7\\x^2=25\end{cases}}\)
=> x ko có giá trị vì x2 \(\ge\)0 mà x2= -7
hoặc x = \(\pm\)5
Ta có \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)(đk : \(x\ne0\))
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
=> \(\frac{7}{x}=\frac{7}{15}\)
=> x = 15 (tm)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
=> \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)
=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
=> \(\frac{1}{3}-\frac{1}{n+3}=\frac{10}{31}\)
=> \(\frac{1}{2x+3}=\frac{1}{93}\)
=> 2x + 3 = 93
=> 2x = 90
=> x = 45
a, đk x khác 0
<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)
b, <=> 36x +252 = -360 <=> x = -17
c. đk x khác -1
<=> (x+1)^2 = 16
TH1 : x + 1 = 4 <=> x = 3 (tm)
TH2 : x + 1 = -4 <=> x = -5 (tm)
d, đk x khác 1/2
<=> (2x-1)^2 = 81
TH1 : 2x - 1 = 9 <=> x = 5 (tm)
TH2 : 2x - 1 = -9 <=> x = -4 (tm)
a: \(\Leftrightarrow x^2=16\)
hay \(x\in\left\{4;-4\right\}\)
b: =>x+7/15=-2/3
=>x+7=-10
hay x=-17
c: \(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)
hay \(x\in\left\{3;-5\right\}\)
a: Để A nguyên thì 2 chia hết cho x
=>\(x\in\left\{1;-1;2;-2\right\}\)
b: Để B nguyên thì \(1-x\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;2;-2;4\right\}\)
c: C nguyên thì \(2x+7\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-3;-4;-1;-6\right\}\)
d: D nguyên
=>x+1+1 chia hết cho x+1
=>\(x+1\in\left\{1;-1\right\}\)
=>\(x\in\left\{0;-2\right\}\)
e: E nguyên
=>x-1+5 chia hết cho x-1
=>\(x-1\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{2;0;6;-4\right\}\)
f: G nguyên
=>2x+6 chia hết cho 2x-1
=>2x-1+7 chia hết cho 2x-1
=>\(2x-1\in\left\{1;-1;7;-7\right\}\)
=>\(x\in\left\{1;0;4;-3\right\}\)
h: H nguyên
=>11x+22-37 chia hết cho x+2
=>\(x+2\in\left\{1;-1;37;-37\right\}\)
=>\(x\in\left\{-1;-3;35;-39\right\}\)
1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)
=>\(5^{x-2}=16+9=25\)
=>x-2=2
=>x=4
2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)
=>3^x=0
=>x=0
3: \(\Leftrightarrow2^x+2^x\cdot16=272\)
=>2^x*17=272
=>2^x=16
=>x=4
4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)
=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)
=>2^x-1=8
=>x-1=3
=>x=4
\(a)\left(2x-1\right)^2+1=26\)
\(\left(2x-1\right)^2=25\)
\(TH1:2x-1=5\)
\(2x=6\)
\(x=3\)
\(TH2:2x-1=-5\)
\(2x=-4\)
\(x=-2\)
Vậy........
\(b)\left(2x-4\right)^3+2=66\)
\(\left(2x-4\right)^3=64=4^3\)
\(2x-4=4\)
\(2x=8\)
\(x=4\)
\(c)7^x+2+5.7^x+1+15=603\)
\(7^x\left(1+5\right)=603-15-1-2\)
\(7^x.6=585\)
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