Đại số: miền xác định, hệ phương trình, bất đẳng thức và hàm số
GRE General Test Chủ đề 12 2:43 Giọng đọc tiếng Anh · phụ đề tiếng Anh + 中文 được ghi trực tiếp
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This unit solves equations while keeping denominator and radical restrictions, analyses inequalities and simultaneous conditions without losing endpoints, and evaluates functions and translates words into algebra.
本单元在保留分母和根式限制的前提下解方程，分析不等式和联立条件而不丢失端点，并对函数求值、把文字关系翻译成代数式。
Write restrictions first: a denominator must be nonzero and a real even root needs a nonnegative argument.
先写出限制条件：分母不能为零，实数偶次根式的被开方数须非负。
Squaring can introduce candidates, so verify them in the original.
平方可能引入候选值，所以要代回原式检验。
In inequalities, a negative multiplier reverses direction; for a factored quadratic, order the roots and test signs.
不等式中乘以负数要变号；对已分解的二次式，把根排序并检验各区间的符号。
A system's solution is a pair satisfying every relation: linear systems have one, none or infinitely many.
方程组的解是满足每个关系的一对数：线性方程组有一个、无解或无穷多个解。
In a composition, evaluate the inner function first.
复合函数先算内层函数。
Read a slope as change per unit, and keep all allowed values in quantitative comparison.
斜率读作每单位的变化量，数量比较中要保留所有允许的取值。
For root x plus 2 equals x, squaring gives 2 and minus 1, but only 2 works.
对 √(x + 2) = x，平方得 2 和 −1，但只有 2 成立。
The upward quadratic x minus 1 times x minus 4 is nonpositive on 1 to 4, including roots.
开口向上的 (x − 1)(x − 4) 在 上非正，包括两根。
Adding x+y=10 and x-y=2 gives x=6 and y=4.
把 x + y = 10 与 x − y = 2 相加得 x = 6、y = 4。
And f of g of 3 is 19.
f(g(3)) = 19。
A transformed equation supplies candidates, not automatic original solutions.
变形后的方程只提供候选值，不一定是原方程的解。
Negative inequality multipliers and strict endpoints need explicit checks.
不等式中的负乘数和严格端点都需要明确检查。
The rational equation excludes 2; for allowed x it becomes x plus 2 equals 5, so x is 3, which checks.
这个分式方程排除 2；对允许的 x，化为 x + 2 = 5，所以 x = 3，经检验成立。
The quadratic is nonpositive between its roots, minus 1 to 3.
二次式在两根之间非正，即 −1 到 3。
The composition applies g first: g of 2 is minus 1 and f of that is 1.
复合函数先算 g：g(2) = −1，f(−1) = 1。
The system x+y=10, x-y=2 meets where x=6 and y=4.
方程组 x + y = 10、x − y = 2 在 x = 6、y = 4 处相交。
Subtract twice the first equation.
减去第一个方程的两倍。
With 8, a minus 2 times y is zero: for a not 2, y is 0 and x is 4; for a equal to 2 the equations coincide, giving every pair 4 minus t, t.
右端为 8 时，(a − 2)y = 0：a ≠ 2 时 y = 0、x = 4；a = 2 时两方程相同，所有 (4 − t, t) 都是解。
With 9, a minus 2 times y is 1: unique for a not 2, and no solution when a is 2.
右端为 9 时，(a − 2)y = 1：a ≠ 2 时有唯一解，a = 2 时无解。
Check yourself.
自我检测。
The radical equation gives 2.
根式方程得 2。
The inequality holds from 1 to 4.
不等式在 1 到 4 上成立。
The rational equation gives 3.
分式方程得 3。
f of g of 2 is 1.
f(g(2)) = 1。
And with a equal to 2 and right side 9, there is no solution.
a = 2、右端为 9 时无解。